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

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.

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

So the trick is to know the interest compounding formula and set it equal to V.

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?

\(v = p (1 + \frac{r}{100})^n - q\)

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

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

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

\(n = 2\)

so, the value increased for two days and the investor sold the third day, the same day when the price fell.
ans: b

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

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"

Apologies for posting a screenshot.I tried to write everything down in comments but messed up the formulas.

Please explain, how we decided to use Compound interest formula?
Thanks in advance

Posted from my mobile device

[quote="shreya1018"]Please explain, how we decided to use Compound interest formula?
Thanks in advance

shreya1018 : I am happy to respond, We have to use CI because in the question it mentioned "For a certain number of days, the value of the share increased by r percent per day"

Everyday it was increasing by r%

So initially if the investor had P.
1st Day : P + r% of P
2nd Day : (P + r% of P) + r% of ( P + r% of P) and so on...

This is exactly what Compound Interest is.So,we used CI formula and not SI.

VeritasKarishma plz explain

Am I correct to say that the answer is easily given from the formula? If it would have been 3 days of growth, then the formula would have been with the cube root and not the sqr root? Five days of growth would give us the fifth root.

Then we can just look at the formula and instantly see there was 2 days of growth here.

Yes, exactly!

the time efficient way to solve this is by using smart number
let p =100 v=300 q=100
(note while choosing smart numbers keep in mind the r should be an positive integer for the ease of calculation )

the use the expression of r which is given
r = 100(sqrt (300+100/100 -1))
r= 100(sqrt (400/100 -1))
r =100
then increase p =100 by 100%
day 1 200 day 2 400 day 3 loss of q$ which is 100 sell will give 300 which is value assumed by us
so
answer is b

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