Solution:

Let us first calculate the total number of matches that will be placed:
9C2 * 2
= 9! / (2!7!) * 2
= 8 * 9
= 72
Therefore, every team will play 16 matches.

Let us now glance over the cases that are possible.

Case 1:
Let us say that Team A, Team B and Team C wins 14 matches each (they lose 1 match each from the other two teams), then the Team D can win only (16 - 3 * 2) = 10 matches.

Case 2:
Suppose that, the top 4 teams win 13 matches each.
Team A will win 13 matches and will lose 3 matches from Team B, Team C and Team D.
Team B will win 13 matches and will lose 3 matches from Team A, Team C and Team D.
Team C will win 13 matches and will lose 3 matches from Team A, Team B and Team D.
Team D will win 13 matches and will lose 3 matches from Team A, Team B and Team C.

If this case actually happens, then the Team E will be able to win 8 matches at maximum as Team A, Team B, Team C and Team D will win all the matches played by E.

Case 3:
Suppose that, the top 5 teams win 12 matches each.
Team A will win 12 matches and will lose 3 matches from Team B, Team C, Team D and Team E.
Team B will win 12 matches and will lose 3 matches from Team A, Team C, Team D and Team E.
Team C will win 12 matches and will lose 3 matches from Team A, Team B, Team D and Team E.
Team D will win 12 matches and will lose 3 matches from Team A, Team B, Team C and Team E.
Team E will win 12 matches and will lose 3 matches from Team A, Team B, Team C and Team D.

If this case actually happens, Team E will win 12 matches but it still won't be able to get into the playoffs assuming that it has the lowest run rate among all.

This implies that from case 2, it is evident that a team must win 13 matches at the least to confirm its position in the playoffs.

Solution:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.

Solution:

This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.

Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.

Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.

Solution:

1 2 3 4 5 6 and 7 8 6

Since summation of numbers 1-9(1 2 3 4 5 6 7 8 9) is 45 , so its impossible to divide into two equal parts.
Since we cant flip over 6 to make it 9 but we can flip 9 to make it 6 and hence the solution

Solution:

199 - 99

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

Flyball should pay Lavy $11 instead of accepting his counter offer as in that case the club will end up paying almost double the amount.

Solution:

The woman told him to take out one nut each from the rest of the wheels and use them.

Solution:

TIF56

Explanation:
The password is formed as:
T (5th letter of Shruti)
I (5th letter of Arshika)
F(1st letter of Friday)
5(5th day of week)
6(Length of Friday)

Similarly, we will pick the 1st letter for Monday and 3rd letter for Wednesday.

Solution:

This is more of a philosophical question than being a riddle or a puzzle. The action of Brian directly led to the result which is the death of Chris. Thus he murdered Chris. In a sense, Chris died due to the lack of water. It is the circumstances that ultimately led to his death.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.