@vbhvgupta said:Find max value of n such that570*60*30*90*100*500*700*343*720*81 is perfectly divisible by 30^n
n = 11
the expression has max power of 3 = 11
max power of 10 = 12..
so it will be divisible by 30^11??
@vbhvgupta said:yes...can you explain?
i will assume all the answers are correct and not just any1 of those....
i solved that question in the hard way.....found out the no. of 5s and 2s in every single term, added the powers and quoted the least among them as the answer.....
i solved that question in the hard way.....found out the no. of 5s and 2s in every single term, added the powers and quoted the least among them as the answer.....
@soumitrabengeri said:What is the OA?
why are u asking this?
in these optionsss
tell me which one is correct:)
@pari19O1 said:Bhai please solve the following question...and post ur solutions...and do tag me...thanks in advance..XAT 2005 question
put (a , b)= 0 , x = 1 , u got -1/3 , eliminate option b..
now put x= 3 , u got 1/5 , eliminate option a.. and option "d" seems incomplete..
so ans is C..
now put x= 3 , u got 1/5 , eliminate option a.. and option "d" seems incomplete..
so ans is C..
@dragster said:put (a , b)= 0 , x = 1 , u got -1/3 , eliminate option b..now put x= 3 , u got 1/5 , eliminate option a.. and option "d" seems incomplete..so ans is C..
aur koi tareeka nhi hai kya??
:|
@vbhvgupta said:yes...can you explain any one of these?
check power of 5.. 250 is wrong calculate sum this series(sum of powers) --> 5 + 10 + 15 +25 till 45 , u get 25*9= 225..
@hari_bang said:why are u asking this?in these optionssstell me which one is correct
How would i know which one is correct?
You posted the question..so you should have the OA
@soumitrabengeri said:How would i know which one is correct?You posted the question..so you should have the OA
i got the ans 13.33
bt ans was 12:)
I dint knw...how?
@gautam22 said:max value 1 easily nikal jayegi aur vo bas ek hi option mein hai to C ans?
yaar ye toh value rakh k pata chal rha hai....
ye toh theek hai....
koi method jaise bachpan mein maths mein karte the...waisi waali chahiye....
:)
@dragster said:may be , koi batao yar conventional method..??
@dragster said:check power of 5.. 250 is wrong calculate sum this series(sum of powers) --> 5 + 10 + 15 +25 till 45 , u get 25*9= 225..
for 25^25, u need to consider two 25s(not 1).......since 25^25 = 5^50...so an addnl 25 will be added to the 25 u found...so 250
@hari_bang said:i got the ans 13.33bt ans was 12I dint knw...how?
I have already explained in the solution posted by me as to how i got 12 as the answer
But i agree that 13.33 is also a possible answer
@dragster said:check power of 5.. 250 is wrong calculate sum this series(sum of powers) --> 5 + 10 + 15 +25 till 45 , u get 25*9= 225..
Ans is 250
@shadowwarrior said:for 25^25, u need to consider two 25s(not 1).......since 25^25 = 5^50...so an addnl 25 will be added to the 25 u found...so 250
yup rite 
@soumitrabengeri said:I have already explained in the solution posted by me as to how i got 12 as the answerBut i agree that 13.33 is also a possible answer
k...thnx..bro:)
@pari19O1
Let, (x^2 - 2x + a^2 + b^2)/(x^2 + 2x + a^2 + b^2) = k
Then, x^2(1-k) - 2x(1+k) +(1-k)(a^2 +b^2)= 0
The above eqn is a Quadratic Eqn in terms of x..Thus, its Discrimant(D) should be >=0
=>b^2 - 4ac>=0
=>4(1+k)^2 - 4(1-k)^2*(a^2 + b^2) >=0
=> (1+k)/(1-k) >= (a^2 + b^2)^(1/2)
=>k>= [(a^2 + b^2)^(1/2) - 1]/[(a^2 + b^2)^(1/2) +1]
Thus, Option C is our Correct Answer... 
gunie bhai logo , vry few noctrnals here 2day
@pari19O1 said:yaar ye toh value rakh k pata chal rha hai....ye toh theek hai....koi method jaise bachpan mein maths mein karte the...waisi waali chahiye....
that a^2 + b^2 is too big for me to type every time...so i will call it 'k' and the whole expr to be y....
so x^2 - 2x + k = y*x^2 + 2x*y + y*k
==> x^2(1-y) + x(-2-2y) + (k-yk) = 0
i will assume that x is real ==> so det >= 0
==> 4(1+y)^2 - 4k(1-y)^2 >= 0
==> k so [ (1+y)/(1-y) ] > rt(k) or [ (1+y)/(1-y) ] with this u can say one limit is [(rt(k) + 1)/(rt(k) - 1)]
and the other is [(rt(k) - 1)/(rt(k) + 1)]
i am also not satisfied with the last line :\
so x^2 - 2x + k = y*x^2 + 2x*y + y*k
==> x^2(1-y) + x(-2-2y) + (k-yk) = 0
i will assume that x is real ==> so det >= 0
==> 4(1+y)^2 - 4k(1-y)^2 >= 0
==> k so [ (1+y)/(1-y) ] > rt(k) or [ (1+y)/(1-y) ] with this u can say one limit is [(rt(k) + 1)/(rt(k) - 1)]
and the other is [(rt(k) - 1)/(rt(k) + 1)]
i am also not satisfied with the last line :\
@pari19O1 said:Bhai please solve the following question...and post ur solutions...and do tag me...thanks in advance..XAT 2005 question
@krum bhaaya krum....krum bhaaya krum....
jab kahin pe kuch nhi bhi nhi tha...tu hi tha tu hi tha...tu hi tha tu hi tha....
:|
@pari19O1 said:Bhai please solve the following question...and post ur solutions...and do tag me...thanks in advance..XAT 2005 question
let x^2 - 2x + a^2 + b^2 / ( x^2+2x+a^2+b^2 ) = P
we get a quadratic : x^2 (P-1) + 2X (P+1) + ( a^2 + b^2 ) ( P-1 ) = 0
at least 1 root will be real & +ve;
b^2 - 4ac > 0 ;
(P + 1) / (P-1) = _/ (a^2+b^2)
P >= _/( a^2 + b^2 ) - 1 / _/ (a^2+b^2) +1
OPTION D ?
wats d OA ?