PG Premier League - Chennai Dream Team

[exile]

Originally Posted by maverick_srikan

Why can't we have a knock out tournament with an odd number of players? 2n+1, with the odd player getting a bye in every round. A knockout tournament with an odd number of players is very much possible.

Yeah, its very much possible.

@Nishant, I did mean 512 matches and not 512 players.

In every match, one team gets eliminated. Therefore, there are 513 team in all.

513 > 2^9.

Ans is 10 rounds.

[nishant_rungta]

Originally Posted by maverick_srikan

Why can't we have a knock out tournament with an odd number of players? 2n+1, with the odd player getting a bye in every round. A knockout tournament with an odd number of players is very much possible.

Well I always thought the whole point with a knock-out tournament is that u cannot have an odd number of players

It's not a knock-tournament in that case. Why would u give a person a bye so that he stands a greater chance to qualify for the finals??

[krsh.vik]

Originally Posted by maverick_srikan

Why can't we have a knock out tournament with an odd number of players? 2n+1, with the odd player getting a bye in every round. A knockout tournament with an odd number of players is very much possible.

Look at this
Single-elimination tournament - Wikipedia, the free encyclopedia

This should clear all doubts if at all there are any

Excerpts from the link above

In cases where the number of competitive entities at the start of the tournament is not a power of two, some competitors may receive a bye in the first round, which entitles these competitors to advance to the second round automatically without playing. Often, these byes will be awarded to the highest-rated competitors in the event as a reward for some previous accomplishment; indeed, in some American team sports - most notably football - the number of teams qualifying for the postseason tournament will be intentionally set at a number which is not a power of two, in order to provide such an advantage to a high-achieving team in the just-completed regular season.
Multiple rounds of byes are also possible: in the FA Cup, the teams in the top two league divisions enter in the third round "proper" (of eight); the two next-highest divisions' teams will have entered in the first round; lower-division teams in one of 6 preliminary rounds.

[satanica]

Originally Posted by exile

...
2) The sum of sqaures of three consecutive odd numbers is a four-digit number N, in which all four digits are same. Find N.
...

Let the three numbers be 2n-1, 2n+1, 2n+3 and the required answer be X.

Sum of their squares = 12(n^2)+12n+11 = X
Hence, X+1 is divisible by 12.

X=5555.

[adityacooool]

Originally Posted by exile

Yeah, its very much possible.

@Nishant, I did mean 512 matches and not 512 players.

In every match, one team gets eliminated. Therefore, there are 513 team in all.

513 > 2^9.

Ans is 10 rounds.

The next time i try a question, i'll first try and imagine all the hidden clauses and interpretations of the questions .

on a positive side, I'll get more type of questions to practise that ways

[ramaan]

Can any one explain this problem came in 0909...

1)How many ordered pairs of real numbers(x,y) satisfy the equation x^3*y^2=864 and x/y(x+y^4)=66 ?

The time solution gave an answer as 2..i got one pair (6,2)..can any one help me out with another pair..????

2)If x,y,z are positive real numbers,for which ratio of the values of y and z is the value of (x/y+z/12x+4y/x+x/3z) the minimum ?

1)1:4 2)1:2 3)1:1 4)2:1 5)4:1

Its again a problem from 0909..

[ramaan]

Originally Posted by exile

2) A set has numbers from 1 to 143. Find the maximum no. of elements in a group that will give the same remainder when divided by 13. ( i hope i'm clear.)

Ans is 12 elements..

the set is (0,13,26,....143) gives the remainder 0.

the remaining set will be (1,14,27,..131) will contain only 11 elements...gives the remainder 1..

same for (2,14,...132)..

similarly every set will have 11 elements except the set which gives remaionder 0 which contains 12 elements...


PS :OOPs sorry didnt read from 1 to 143..so the answer is 11...

[an_elusive_lady]

Originally Posted by ramaan

Can any one explain this problem came in 0909...

1)How many ordered pairs of real numbers(x,y) satisfy the equation x^3*y^2=864 and x/y(x+y^4)=66 ?

The time solution gave an answer as 2..i got one pair (6,2)..can any one help me out with another pair..????

2)If x,y,z are positive real numbers,for which ratio of the values of y and z is the value of (x/y+z/12x+4y/x+x/3z) the minimum ?

1)1:4 2)1:2 3)1:1 4)2:1 5)4:1

Its again a problem from 0909..

for the second question, use the logic that if the product of x and y is constant, then the sum is minimum when x=y. if u consider x/y * 4y/x , product is 4, hence sum is minimum when x/y = 4y/x = 2. similarly, z/12x * x/3z = 1/36 => 3z/x = 6 => z/x = 2.
now z/x * x/y = z/y = 4. hence answer is y:z = 1: 4

hope this helps...

also @ ramaan, belated happy bday. cdnt wish u yest as pg was not allowing me to log in

[monsterkartik]

Originally Posted by ramaan

Can any one explain this problem came in 0909...

1)How many ordered pairs of real numbers(x,y) satisfy the equation x^3*y^2=864 and x/y(x+y^4)=66 ?

I am not able to get the question

whether the second equation is x/y*(x+y^4+66 or x/y(x+y^4) = 66 ??

as in whether (x+y^4) is in the denominator or numberator ??


Cheers

MK

[maverick_srikan]

Originally Posted by ramaan

1)How many ordered pairs of real numbers(x,y) satisfy the equation x^3*y^2=864 and x/y(x+y^4)=66 ?


Its again a problem from 0909..

There is just 1 solution to this, and i guess TIME also gave the solution as 1. Can you pls check and confirm?

x^2/y+y^3*x=66

let the two terms be a and b.

Now a*b=864 and a+b=66. We get just 1 solution, if I am correct!