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Quantitative questions a day 51 to 100 - The discussions Quantitative

I'm starting a new thread as the discussions will happen on this new thread for next 50 questions and the sectional test :-)
Happy solving :-)
Thanks a lot yaar, i did a very stupid error.Vaise we are on the wrong thread:) see you on the other one!!
hi guys ..this is my first post at PG ...
My answer is also 20/3
me also getting as option a - 20/3
Shouldnt ur first eqn be 9 = (18/x-v) - (18/x+v)? Because x-v is going to be smaller than x+v, so 18/x-v > 18/x+v
18/(x-y)-18/(x+y) =9 =>2y/x^2-y^2=1/2(x spped of rowin,y speed of stream)
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=...
Question 101:
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
ONE HELP
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..
Abhi mil gaya na answer, m=-n^2...But, its lil long..let's see if we can find some different approach...
mpq+nq=1
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...
yeah ur right...
Whatever the 2nd term be, it isnt zero.....
actually stopped solving when i saw (m+n)
hi me too getting the same answer, i approached with some steps like taking p=q=r=1 and then from the options was able to get the option 3). One help guys please..I joined late in this forum so was able to get questions only from #89..If anyone could forward the previous questions to this id:jps...
@koolbuddy
u r right.... even im getting the it as (m+n^2)( mp^2 +np-1) = 0
Hey Mansoor...
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)=q, f(q)=r, f(r)=p => f(f(f(p)))=p
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.
=>...
Even I did that ... going by answer options after getting an eqn in terms of p, m and n.
But could not get LHS = RHS by any of the answer options......
I suppose the answer is c. After working on equations i checked from options which relation can be possible.
Same here, not able to reach a solution.