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:

1st Bag = One Coin
2nd Bag = Two Coins
3rd Bag = Four Coins
4th Bag = Eight Coins
5th Bag = Sixteen Coins
6th Bag = Thirty Two Coins
7th Bag = Sixty Four Coins
8th Bag = One Hundred and Twenty Eight Coins
9th Bag = Two Hundred and Fifty Six Coins
10th Bag = Remaining Coins.

Now imagine any amount in your mind and you will find out that you can give any amount by just using the pouches (without opening them)

Solution:

Assume that P, Q and R are the three friends. Now P is earning X amount, Q is earning Y amount and R is earning Z amount.

First, P will add his salary to a random amount A and will tell the combined sum to Q.
X + A

Now, Q will add his salary and a random amount B to the amount passed by P. He will then pass it to R.
X + A + Y + B

Now R will add his salary and a random amount C to the total amount passed by Q.
X + A + Y + B + Z + C.

Now, R will pass this amount back to P. P will deduct the random amount added by him in the beginning.
X + Y + B + Z + C

He will now pass this amount to Q who will subtract the random amount he added in his turn.
X + Y + Z + C

He will pass this amount to R who will deduct his random amount.
X + Y + Z

Now this is the total sum of their salaries. R will tell it to P and Q and then each of them can find the average of their salaries by dividing the total amount with 3.

Solution:

Pick from the box that is labelled as 'chocolates and toffees’. Since it is an incorrect label, it must have either a chocolate or a toffee.

Suppose if you find a chocolate, then label it as chocolate. Now change the box that was labelled as chocolate earlier and label it as toffees and the box labelled as toffees to 'chocolates and toffees. This is because the one that was labelled as toffees (since it is labelled incorrectly) must have either chocolates or both chocolates and toffees. Since we have already found the box that has chocolates, it must have both chocolates and toffees.

Solution:

Let us denote the four people with A, B, C and D.
A takes 1 minute to cross, B takes 2, C takes 5 and D takes 10.

A and B cross first spending 2 minutes.
A comes back with torch taking 1 minute.
C and D cross taking 10 minutes.
B comes back with torch taking 2 minutes.
Finally, A and B cross the bridge taking 2 minutes.

2 + 1 + 10 + 2 + 2 = 17 minutes

Thus, this is the way they all managed to cross that bridge that night.

Solution:

Since you can't identify which tablets are meant for which purpose, what you can do is take half of each tablet. In this manner, you will be able to get 1 of each tablet no matter what.

Solution:

The best way is to divide the batteries in three groups. Take three batteries in two groups each and two batteries in the third group.

In the groups with three batteries, you can have three possible combinations in each group. In the group with two batteries, you can have just one possible combination.

If you add the combinations: 3 + 3 + 1 = 7, you get the least number of pairs you must check.

Solution:

Here, there are two events that are independent of each other. Now, in such case, we calculate the probability of both occurring A and Bas:
P(A and B) = P(A) * P(B)
P(Pink) = 1/5
P(Pink and Pink) = P(Pink) * P(Pink)
= 1/5*1/5
= 1/25

Solution:

The total population of New York City is about nine millions. Let us assume that 1 million among them are bald which leaves us with eight millions of people that are required to have different number of hairs on their head.

Now as per the surveys, on an average, people have 100, 000 hairs on their head. Since all must have different numbers, we start with one. Let us say someone has one hair strand, someone has two, and someone has three and so on. After 100, 000 people, we are left with 7, 900, 000 people left. Since all of them are needed to have different number of hair strands, we will move up from 100, 000.

In such case, at the end, you will be having a person with 8, 000, 000 hairs which is clearly impossible (even having 1, 000, 000 hairs on someone's head is impossible).

Therefore, there must be two people who have same number of hairs on their head.

This is a classic example of the pigeonhole principle.

Solution:

The first step will be to cut the three tablets in equal halves and then put each half in two piles. Now, each of the two piles contains half of pill A and two halves of pill B (or one pill B). Now, take out one pill A from the bottle and cut it into half. Put the two halves in two piles.

In this way, each pile now contains two halves of pill A and two halves of pill B or 1 pill A and 1 pill B. You can now take 1 pile one day and the second pile on the next day.