stocks

ddm · Posted: Fri Aug 29, 2008 3:32 pm Post subject: stocks

attachment

parallel_chase · Posted: Fri Aug 29, 2008 4:41 pm Post subject:

stock bought = $p

increasing rate = r/100

decreased by=$q

sold for = $v

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

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

(r/100)^2 = (v+q)/p - 1 = [v+q - p]/p

p* (r/100)^2 = v+q -p

p + p(r/100)^2 = v+q

1st day increase = r/100
2nd day increase = r/100

total increase = (r/100)^2

total price of stock on 2nd day = p + p(r/100)^2

3rd day stock decreased by $q
sold the stock for $v

stock price - $q = $v

Therefore,

p + p(r/100)^2 - q = v

p + p(r/100)^2 = v+q

how many days after buying the share stock was sold

1st day increase
2nd day increase
3rd day decrease

Total 3 working days.

Took sometime to do this problem. If you dismantle the equation given in the question, this problem can be done in less than 1 minute. Let me know if you have any doubts.

ddm · Posted: Fri Aug 29, 2008 6:23 pm Post subject:

thanks a lot...

Fab · Posted: Sun Aug 31, 2008 11:00 am Post subject:

To parallel chase:

How did you get from:

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

(r/100)^2 = (v+q)/p - 1 = [v+q - p]/p

I thought [(v+q)/p - 1] ^2is a square binomiun...

THANKS

sudhir3127 · Posted: Sun Aug 31, 2008 11:15 am Post subject:

Fab · Posted: Sun Aug 31, 2008 12:35 pm Post subject:

Hello Sudhir, i'm sorry if i'm missing something up, but isn't:

[sqrt[(v+q)/p] - 1 ]^2 = v+q/p - 2sqrt v+q/p + 1 ??

THANKS

sudhir3127 · Posted: Sun Aug 31, 2008 10:10 pm Post subject:

mals24 · Posted: Mon Sep 01, 2008 4:17 am Post subject:

sorry sudhir i didn't get ure explanation i have the same doubt as FAB

In the question only (v+q)/p is under the root and 1 isn't so sqrt[(v+q/p] is one term and 1 is another term so when we square it shouldn't we use the formula (a+b)^2 where a=sqrt[(v+q/p] and b=1

squaring it we should get (v+q)/p + 1 - 2sqrt[(v+q/p]

or is 1 also under the root because if it is then your method would be right but its a separate term outside the root.

hengirl03 · Posted: Sat Sep 06, 2008 4:29 pm Post subject:

When you guys see a problem like this one, what is the first thing that you think to do?

cramya · Posted: Sat Sep 06, 2008 7:06 pm Post subject:

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

We are give the initial stock price is p and it increases by r% for a certain number of days

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

n above is the no of days for which the stock increases by r % Sp the number of days after which stock was sold would be n + 1
(since its given the stock was sold the next day after the number of days the strock increased by r%)

Coming back to this equation p (1+r/100) ^ n - q = v

Since square root was involved I took the educated guess of plugging in 2
for n

so p (1+r/100) ^ 2- q = v

(1+r/100) ^ 2 = v+q/p

Taking sqareroot on both sides

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

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

Therefore n = 2 and n+1 = 2+ 1=3. The stock was sold after(keword is AFTER) 3 days (includes the day stock was sold)

cramya · Posted: Sat Sep 06, 2008 7:09 pm Post subject:

I also agree with Mals24 and Fab's question.

The 1 is not inside the square root. Its outside the squareroot

so its not sqrt ((v+q/p) - 1) its

sqrt(v+q/p) - 1

karthikgmat · Posted: Sat Sep 06, 2008 10:09 pm Post subject: I think its just two days

let me follow like this
cost = $p
profit = r/100
so total price after 1day = p[1+r/100]
for the second day it will be p * ([1+r/100]^2)
and so on like that....
on the nth day the price of it will be p*([1+r/100]^n)

fall in price of stock is $q

his selling price = $v so

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

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

from the give eq

r = 100 (sqrt ((v+q) / p) - 1) and sub v+q)/p= [1+r/100]^n
we get n=2 i think its two days..
Am i Right..