We are told that there are six teams, that are divided into two groups.
Teams in the same group will play each other only once, and teams in different group will play each other twice.
Calculating the combinations, there is going to be games among teams in the same group among them,
and since there is two groups, total such games will be 6.
Now, teams in different group play each other twice. Calculating the combinations for this,
From the first group, a team can be chosen in three ways, and from the second group, a team can be chosen in
three ways. Total ways two teams from different groups can play each other is 3x3 which is 9. And since they
play each other twice, that is 9+9 games of this combination.
Total number of games is 18+6=24
It is given that each team plays one game in each round, that means there is going to be 3 matchups in each
round. And given, there is 24 games to played in this format, the number of rounds will be 24/3=8
The tournament will have 8 rounds.
Now that we know there are going to be 8 rounds in the tournament, let us identify the teams in a particular
group, that will help us build the matchups.
We are told that Round 8 teams from different groups play each other, and Teams 1 and 5 play only once. This
means, 1 and 5 have to be on the same group. It is also told that 4 and 6 play each other twice, that means 4
and 6 have to be in different groups. Looking at the matches from Round 8 that is given to us, 3 played 6 and 2
played 5. We already know 1 and 5 are in the same group, so 2 must be in the other group. Among 3 and 6, if we
were to place 3 in the group with 1 and 5, 4 and 6 would have to be in the same group, which is not possible,
hence 6 is with 1 and 5, giving us the final combination of groups.
Now, using the given information to build the matchups for the 8 rounds.

Rounds marked with the same colour represent the fact that the matchups are identical. Now we know that
each team plays a game in each round, we know 2 out of the 3 matches for Round 2 and 8, and we can identify
the third matchups as well. Giving us this resulting table.
We are told that Round 4 and 7 are identical, that means they are the matchups between teams from two
different groups,
We look at the matchups that are remaining among the 6 teams where both the games are left to play.
Right away we identify that 6 is yet to play 2 twice and 5 once. We are looking for teams playing twice, so both
Round 4 and 7 has a matchups between 2 and 6. This means 6 will play 5 in Round 3 and using that we can
identify the third matchup in Round 3 as well.
Now, we can identify that 1 is yet to play 3 both the times, 2 once. And we are looking for teams playing each
other twice.

There are two cases formed since we cannot identify for certain two whom did O reward the stars to,
We can identify the number of stars rewarded by M to C.
Hence only statement 1 can be identified uniquely.

M distributed among 3 bloggers, N among 2 bloggers, O among 1, P among 2, X among 1, Y among 3
Hence 2 surfers distribute their stars among 2 bloggers.

We are told that D receives more stars than C from Y. Considering Y has already given 25 stars, it will give 0
stars to C and 5 stars to D.
The only two surfers who have not given any stars to A or B is O and X, and these are the two surfers to give all
of their stars to a single blogger.
We are also told that M gives different stars to the four bloggers,
Since he has already given 0 and 10, the remaining distinct stars should add up to 20. The only numbers that
are remaining that add up to 20 are 5 and 15.
We know that X rewards one of C or D 30 stars and O rewards one of C or D 30 stars. Given that, M could not
have rewarded D 15 stars, since Y rewarded D 5 stars, and D is also going to rewarded 30 stars by O or X, and
since the total is same for all, which is 45. This is not possible.
This means that, M rewarded C 15 stars and D 5 stars.
This gives us two cases

We are told that D receives more stars than C from Y. Considering Y has already given 25 stars, it will give 0
stars to C and 5 stars to D.
The only two surfers who have not given any stars to A or B is O and X, and these are the two surfers to give all
of their stars to a single blogger.
We are also told that M gives different stars to the four bloggers,
Since he has already given 0 and 10, the remaining distinct stars should add up to 20. The only numbers that
are remaining that add up to 20 are 5 and 15.
We know that X rewards one of C or D 30 stars and O rewards one of C or D 30 stars. Given that, M could not
have rewarded D 15 stars, since Y rewarded D 5 stars, and D is also going to rewarded 30 stars by O or X, and
since the total is same for all, which is 45. This is not possible.
This means that, M rewarded C 15 stars and D 5 stars.

There are three bullish shares, C D and G
Lets say a trader buys one share of each of these stocks, and sells them at their day's high
One share of C at opening is 800, sells at 1400
One share of D at opening is 500, 1200
One share of G at opening is 1200, 1900
Total Investment is 2500, and total money after selling is 4500
That is an 80%.

Daily loss for a share is defined as (Opening price - Closing price) / (Opening price)

Stock G gained money that day
Hence Stock A has the highest Daily Loss.