if p=3*5*7*9*…99/2*4*6*8*…100 then p lies between 1) 1/2 and 1/3 2)1/3 and 1/5 3)1/5 and 1/10 4)1/10 and 1/15 I can do this question by eliminating options but how to do this fundamentally…

if p=3*5*7*9*............99/2*4*6*8*.....................100

then p lies between

1) 1/2 and 1/3

2)1/3 and 1/5

3)1/5 and 1/10

4)1/10 and 1/15

I can do this question by eliminating options but how to do this fundamentally.....

if p=3*5*7*9*............99/2*4*6*8*.....................100

then p lies between

1) 1/2 and 1/3

2)1/3 and 1/5

3)1/5 and 1/10

4)1/10 and 1/15

I can do this question by eliminating options but how to do this fundamentally.....

Hi Whocarez....i did give it a try but not sure if gives an exact enuff result

okay, here goes

eqn can be written as

3.4.5............................99

-------------------------------

2.1 . 2.2 . 2.3 . 2.4 ..... 2.50

3.5.7...............99

---------------------

2^50 (1.2.3....50)

51.53.57.........99

-------------------

2^50(2.4.6.....50)

(1 \2^25).(51\52)......( 99\100)

normalize the equation to contain only the first and the last term, i.e. the lower and upper limits respectively...

i.e eqn will look something like

(1 \2^25).(51\52)...25 times.....( 51\52)

and

(1 \2^25).(99\100)...25 times.....( 99\100)

now, the above eqn have have values ranging from (51/104)^25 to (49.5/100)^25 which i think can be safely approximated to lie between 1/3 to 1/2

what say...??

Vinayak

seems logical but wrong answer :D

just rewrite the series as 1/2*3/4*5/6*.......99/100. now if u multiply the first three terms u get 5/16 which is less than 1/3 followed by 47 fractions each of which is less than 1. So p is less than 1/3 😃 :)

some1 plz give the soln.

i can give one part of the answer.. the other part i still hafta find out.

consider p'=100*p= (3/2)*(5/4)*....(99/9

and q=(2/1)*(4/3)*(6/5)*....(98/97)

we can see tht each term of q>p' and so q>p'

now, q*p'=99

which implies p'^2 ->p=p'/100

right?

tht gives choice 4.....

why greater thn 1/15?

...lemme think

yuhooo.... here goes the second part.

p=1/2*( 3/4)*(5/6).... (99/100)

let p'=2p=(3/4)(5/6)....(99/100)

q=(2/3)(4/5)(....(98/99)

->p'*q=1/50

and p'>q

->p'^2>1/50

->p^2>1/200>1/225

->p>1/15 😃

hi

very gud solution, josh i must say......donno if it is worth spending that much time in a CAT exam though...

actually if u multiply the numerator by 2*4*6*...100. u can write it as 100!/(2^50 *50!^2), and there was some easy formula or inequality for n!/(n/2)!^2....forgot that one.. wud be easier if u cud find that formula... will post it i can find it....

regards,

rakesh

That was brilliant josh!!!! :):)

hey rakesh please find the formula let us know 😃

HIII

this is vidhan also a call from pgdcm& pgdm iim cal how r u preparing help me out

HI Guys ,

I am a newbie here. anyway solution to the question below :

1.if p=3*5*7*9*............99/2*4*6*8*.....................100

then p lies between

1) 1/2 and 1/3

2)1/3 and 1/5

3)1/5 and 1/10

4)1/10 and 1/15

Solution: This is a general solution.Not a short cut /or otherwise.

IT can be used to solve other factorial realted problems.

p = 3*5*7*9....99 / 2*4*6...100

= 100! / (2^50 *2^50*50!*50!)

=100! /(2^100*50!*50!)

Now comes the important formula called Stirling's formula:

n!~ (n/e)^n * sqrt(2*pi*n) (read ~ as approximately)

So usign this formula suubstitute all the values

u shud get something like this

p= 1/(sqrt(50*pi)) = 0.079

so the choice is D.

Note: There is a accurate version of this formula also but i guess this is more than enough.

comments are welcome!

regards,

arun.