then according to the conditions given,a-2,a+d,a+2d+10 will be in gp
(a+d)^2 =(a-2)(a+2d+10)
--->(d+2)^2=8a-20
only d will statisfy this condition
@vbhvgupta a,a+d,a+2d (Assume )
@amitranjan2311 said:@vbhvgupta a,a+d,a+2d (Assume )then according to the conditions given,a-2,a+d,a+2d+10 will be in gp(a+d)^2 =(a-2)(a+2d+10)--->(d+2)^2=8a-20only d will statisfy this condition
(d+2)^2=8a-16 hoga
Q7
@mailtoankit said:4 numbers?
did you get a=d (for the AP)??
i am trying to do it frm office....muje hua ni :(
@vbhvgupta said:Q7
@Logrhythm said:did you get a=d (for the AP)??i am trying to do it frm office....muje hua ni
Here, Hundreth Digit= a+d, Tenth Digit= a n Unit Digit= a-d
Thus, A= (a+d)(2a - d) + a(2a) + (a - d)(2a + d)
=> 6a^2 - 2d^2
, &, B=a(3a)= 3a^2..Thus, 6a^2 - 2d^2= 4a^2, or, a=d..
Thus, Possible value of N are 210, 420, 630 n 840.. Hence Option(A). 4 will be our answer..
@Logrhythm said:did you get a=d (for the AP)??i am trying to do it frm office....muje hua ni
haan...got a=d ...getting 4 numbers 210,420,630,840
@vbhvgupta said:correct..approach bhai??
H........T........U
a+d.....a........a-d
(a+d+a)(a-d) = (2a+d)(a-d)
(a+a-d)(a+d) = (2a-d)(a+d)
(a+d+a-d)(a) = (2a)(a) = 2a^2
adding all these values..will get
6a^2-2d^2=A
a(a+d+a+a-d) = B
3a^2 = B
A=4/3B
6a^2-2d^2 = (4/3)(3a^2)-----> a = d
H......T.......U
2a....a........0
210,420,630,840...satisfies the above condition
@mailtoankit said:haan...got a=d ...getting 4 numbers 210,420,630,840H........T........Ua+d.....a........a-d(a+d+a)(a-d) = (2a+d)(a-d)(a+a-d)(a+d) = (2a-d)(a+d)(a+d+a-d)(a) = (2a)(a) = 2a^2adding all these values..will get6a^2-2d^2=Aa(a+d+a+a-d) = B3a^2 = BA=4/3B6a^2-2d^2 = (4/3)(3a^2)-----> a = dH......T.......U2a....a........0210,420,630,840...satisfies the above condition
lol a=d ke baad mein akal ni chala paaya....boss baitha hai na side mein toh akal kaise chalegi...... :p
In a chess tournament-
evry participant played 2 games wid odr participant.
no f games tat men played among themselves exceeded the no f games tat men played wid women, by 66.
1) no f participant in tournament? 12,13,11,15
2) total no f games played? 132, 110, 156, 210
@The_Loser said:In a chess tournament-evry participant played 2 games wid odr participant.no f games tat men played among themselves exceeded the no f games tat men played wid women, by 66.1) no f participant in tournament? 12,13,11,152) total no f games played? 132, 110, 156, 210
are u sure the question has complete data???
@The_Loser said:In a chess tournament-evry participant played 2 games wid odr participant. no f games tat men played among themselves exceeded the no f games tat men played wid women, by 66.1) no f participant in tournament? 12,13,11,152) total no f games played? 132, 110, 156, 210
=> 2*[mC2 - mC1*nC1] = 66
=> mC2 - m*n = 33
=> m*[(m-1)/2 - n] = 11*3
m can only be 11 and so n = 2.
1.) No. od Participants = 13.
2.) Total Games Played = 2*13C2 = 13*12 = 156.
@The_Loser said:In a chess tournament-evry participant played 2 games wid odr participant. no f games tat men played among themselves exceeded the no f games tat men played wid women, by 66.1) no f participant in tournament? 12,13,11,152) total no f games played? 132, 110, 156, 210
getting ajeeb equation and according to that equation my answer doesnt match with ur given answer either i am wrong or question is
@Estallar12 said:=> 2*[mC2 - mC1*nC1] = 66=> mC2 - m*n = 33=> m*[(m-1)/2 - n] = 11*3m can only be 11 and so n = 2.1.) No. od Participants = 13.2.) Total Games Played = 2*13C2 = 13*12 = 156.
ye equation to tab banega na jab humein kuch pata ho how many mens or womens are there ? Correct me
@shattereddream said:getting ajeeb equation and according to that equation my answer doesnt match with ur given answer either i am wrong or question is
there are 2 women, that should be given in the question
let no. of men=m
m(m-1)-m*4=66
m=11 and no. of games =13*12=156
@shattereddream said:ye equation to tab banega na jab humein kuch pata ho how many mens or womens are there ? Correct me
Nope.
See, Suppose there are m Men and n Women.
Now, Given 66 number can be written in a equation like this :
2*mC2 = No. of Games between Men.
2*mC1*nC1 = No. of Games between Men and Women.
Difference = 66. Equate it.
Then on factorising 33 = 11*3, we can see that m cannot be 3 else, the term in bracket will go negative. So, m = 11 and n = 2.
@junefever said:there are 2 women, that should be given in the questionlet no. of men=mm(m-1)-m*4=66m=11 and no. of games =13*12=156
2 Women is not needed.
We can form the equation and find that out easily.
Question is complete. :)
@junefever said:are u sure the question has complete data???