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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...
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..
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)
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...
(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......