Happy solving :-)

Happy solving :-)

hi guys ..this is my first post at PG ...

My answer is also 20/3

My answer is also 20/3

me also getting as option a - 20/3

18/(2x-y)-18/(2x+y)=1=>2y/4x^2-y^2=1/18

hence 2y=(x^2-y^2)/2 =(4x^2-y^2)/18

from (x^2-y^2)/2 =(4x^2-y^2)/18 we get 5x^2=8Y^2 =>5(x^2-y^2)=3y^2=>x^2-y^2=3/5*y^2

put ths vaue in eqn 2y/x^2-y^2=1/2 u get 10y/3y^2=...

At his normal speed, Shirish can travel 18 km downstream in a fast flowing stream in 9 hours less than what he takes to travel the same distance upstream. The downstream trip would take 1 hour less than what the upstream trip would take provided he doubles his rate of rowing. Wh...

U can get some of the Qs here....

http://pagalguy.com/newsletter/archive.php?ArchiveID=MTE%3D

http://pagalguy.com/newsletter/archive.php?ArchiveID=MTE%3D

guys pls could anyone forward the quatz quetion a day mails before question #89...it will be of gr8 help to me... thnx in advance..

mqr+nr=1

mpr+np=1

i took value of q from 1st eq..

put it in 2nd .

den d value of r frm 2nd eq n put it in 3rd one..

it was a long method.. n made mistakes too...

Whatever the 2nd term be, it isnt zero.....

actually stopped solving when i saw (m+n)

@koolbuddy

u r right.... even im getting the it as (m+n^2)( mp^2 +np-1) = 0

u r right.... even im getting the it as (m+n^2)( mp^2 +np-1) = 0

Ur solution is fine...

but dont u think that the last term shud be...

(m+n^2)( mp^2 +np-1) = 0

That's wat i m gettin...

Though ultimately both leads to...

m=-n^2

got it Mansoor Bond !!

f(p)=1/(mp+n)

f(f(p))=1/(m(f(p)+n)=(mp+n)/(m+n+mnp)

similarly,

f(f(f(p)))=(m+mnp+n)/((m+n)p+n(2m+n))=p

=> (m+n)p+n(m+n)p+(m+n)=0

=>(m+n)(p+pn+1)=0

(p+pn+1) !=0 since similar relations will hold for q, r making p, q, r not distinct.

=>...

But could not get LHS = RHS by any of the answer options......

Same here, not able to reach a solution.