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.
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Source: Beat The GMAT — Problem Solving |
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
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
Me replying for chase,... hope he doesnt mind ...Fab wrote: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
r/100 = [sqrt[(v+q)/p] - 1 ]
u need to do squaring on both sides
it will be
(r/100)^2 =[sqrt[(v+q)/p] - 1 ]^2
square root and whole square get knocked off
we have
(r/100)^2 = (v+q)/p -1
take lcm on rhs.. its p
v+q -1*P/p
= ( v +q - p)/p
hope tha helps.. do let us know if u have any doubts
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
[sqrt[(v+q)/p] - 1 ]^2 = v+q/p - 2sqrt v+q/p + 1 ??
THANKS
Fab wrote: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
U shud see this as
[square root ( a-b)]^2
it will be just a-b but not a+b-2rt ab
i think u r getting confused with
(sqrt a- sqrt b )^2
in this case it will be
(sqrta)^2+ (sqrtb)^2 - 2( sqrta)(sqrtb)
it will be
a + b- 2rt ab
hope this helps ..do let us know if u still have doubts...
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.
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.
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)
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)
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
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
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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..
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..
