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An investor purchased a share of nondividendpaying stock [#permalink]
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14 Apr 2007, 17:55
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An investor purchased a share of nondividendpaying 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.
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Re: Another 700 level Question  Percents [#permalink]
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15 Apr 2007, 23:54
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Witchiegrlie wrote: An investor purchased a share of nondividendpaying 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.
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Re: An investor purchased a share of nondividendpaying stock [#permalink]
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20 Feb 2012, 06:05
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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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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: PSMGMAT CAT 1 [#permalink]
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06 Oct 2007, 06:16
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singh_amit19 wrote: An investor purchased a share of nondividendpaying 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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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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An investor purchased a share of non dividend paying stock [#permalink]
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17 Mar 2011, 05:59
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An investor purchased a share of non dividend paying stock for p dollars on monday.For a certain number of days the value of share increased by r % per day.After this period of constant increase,the value of shARE decreased the next day by q dollars and the investors 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[(v+q)/p]^1/21) A.2 working days later B.2 working days later C.4 working days later D.5 working days later E.6 working days later
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Last edited by AnkitK on 17 Mar 2011, 08:48, edited 1 time in total.



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Re: An investor purchased a share of non dividend paying stock [#permalink]
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17 Mar 2011, 07:38
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r% per day = 365r % per annum = 3.65r Original equation is; \(FV=PV(1+\frac{r}{n})^{nt}\) Here PV=p FV=v r = rate of interest per annum = 365r% = 3.65r n=periods or frequency with which the rate is applied=365(the rate is applied daily; thus making the frequency per year as 365) t= time in years= d days = d/365 years The equation then would look something like this; \(v = p(1+3.65r)^{(\frac{d}{365}*{365})}q\) ### q is the extra bit because after the money value increased for d days; it plummeted by $q on the last day "(d+1st) day". \(v = p(1+\frac{3.65r}{365})^{d}q\) \(v = p(1+\frac{r}{100})^{d}q\) Now, try substituting for r; \(r = 100(\sqrt{\frac{v+q}{p}}1)\) \(v = p[1+\frac{100(\sqrt{\frac{v+q}{p}}1)}{100}]^{d}q\) \(v = p[1+\sqrt{\frac{v+q}{p}}1]^{d}q\) \(v = p(\sqrt{\frac{v+q}{p}})^{d}q\) \(\frac{v+q}{p} = (\sqrt{\frac{v+q}{p}})^{d}\) Let \(\frac{v+q}{p}\) be \(\Delta\) \(\Delta^1 = \sqrt{\Delta}^{d}\) Squaring both sides \(\Delta^2 = \Delta^{d}\) d=2 Thus he sold the stock on d+1 = 2+1 = 3 days AnkitK, could you please correct the question?
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Re: An investor purchased a share of non dividend paying stock [#permalink]
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17 Mar 2011, 20:29
it takes maybe about 2.30 mins to make a count if you know about future value and present value in finacial stuff.... future value is FV as fluke said present value is PV and r is interest for sure the formula is FV=PV(1+r%)^n (n is number of period) here you dont need to divide by 365 days or 360 days, because r in the question is counted by days as the question said ( 360 days or 365 days is decided by banks to count interest..so we dont care about it) . Thus we will have the fomula like
P present value, future value is V and r is know
P*{{1+100*([(r+q)/100p]^1/2)1}^n}q=v ( we have to minus q because after n+1 day the future value will be decreased by q and we have to divide by 100 because its r% ok? <=> p*{(r+q)/p}^1/2}^n=v+q or {[(r+q)/p]^1/2}=(v+q)/P here we see that only n=2 works with the above formular



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Re: An investor purchased a share of non dividend paying stock [#permalink]
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17 Mar 2011, 20:51
Simpler quicker approach. Look at the equation given. v+q is the price before the final drop. Divided by p you have the total multiplier of the stock price's increase. This is the %increase for one 'term' if you subtract 1. But it is ^(1/2) which means you are looking at two terms, and since the question tells us that r% refers to 1 day, then 2 terms is 2 days.
Just remember to add 1 at the end for the day it decreased, and=3.



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Re: An investor purchased a share of non dividend paying stock [#permalink]
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17 Mar 2011, 21:55
Dear Boston,pls explain the meaning of the statement more precisely " But it is ^(1/2) which means you are looking at two terms, and since the question tells us that r% refers to 1 day, then 2 terms is 2 days. " Still not clear
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Re: An investor purchased a share of non dividend paying stock [#permalink]
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17 Mar 2011, 22:43
If the total multiplier from inital price (I) to end price (E) is 'x', i.e. E=Ix, then:
for one period of price changes, the period multiplier (1+%r in this problem) will be equal to the total multiplier, =x.
for two periods, the period multiplier will be the square root of x since the I will be multiplied twice by it to reach E.
for n periods, the period multiplier (1+%r) will be the nth root of x.
I dont like the formula used above re: /365 and per annum etc, because the question's period could have referred to e.g. seconds instead of days and we dont really want to be turning seconds into per annum..



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Re: An investor purchased a share of nondividendpaying stock [#permalink]
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26 Oct 2015, 00:50
Monday: P Tuesday: p(1+r/100) Wednesday: p(1+r/100)^2 Thursday: p(1+r/100)^3
we have square root in the definition of r, so it was the second day of increase + one day of decrease
B



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Re: An investor purchased a share of nondividendpaying stock [#permalink]
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26 Oct 2016, 00:44
I have a small doubt, how do we identify whether we solve it using Simple interest or compound interest



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Re: An investor purchased a share of nondividendpaying stock [#permalink]
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23 Jun 2017, 22:31
Hi: Kindly explain, how we decided to use Compound interest formula? Thanks in advance



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Re: An investor purchased a share of nondividendpaying stock [#permalink]
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24 Jun 2017, 03:18
Witchiegrlie wrote: An investor purchased a share of nondividendpaying 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. Simple we can deconstruct this problem by constructing variables so let p =$2 Q=$12 V=$60 R= 100(\sqrt{60 + 12/2}1_ R= 100(5) R= 500 percent So if the stock was initially bought at $2 dollars on day 1 then it had increased to $12 dollars on day 2 and then 72$ on the third day which is the day it was sold  three days. Thus "B"




Re: An investor purchased a share of nondividendpaying stock
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