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.

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:

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:

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:

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:

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:

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:

This puzzle is quite similar to the cheating husbands puzzle. The same logic applies here.

After the announcement, if there is only one person with green eyes, he will observe everyone and will come to the conclusion that he himself has green eyes.

If two persons have green eyes, they will see everyone and find out that there is one person with green eyes. So they will expect him to leave. But if no one leaves on that day, on the next day both of them will understand that they both have green eyes and will leave the island.

Thus if you keep going like this or apply the rules of induction, you will find that since there are 100 green eyed people on that island, all of them will leave the island on the 100th day after the announcement.

Solution:

12 minutes

Explanation:
Yes, surely they can all make it out of the cave.

First, father and mother will go. Time taken = 2 minutes.
Father comes back. Time taken = 2 + 1 = 3 minutes.
Both children go. Time taken = 3 + 5 = 8 minutes.
Mother comes back. Time taken = 8 + 2 = 10 minutes.
Mother and Father go. Time taken = 10 + 2 = 12 minutes.

Thus in 12 minutes, all of them will be out of the cave through the tunnel.

Solution:

Just form an equilateral triangle, planting trees at the vertices. Now find the center of the triangle and plant the fourth three on that point. In this manner you will form a shape of tetrahedron and all the trees will have equal distance between each other.