Solution:

The package comprises of the parachute which is tied to her back while she is diving free in the air. If she doesn't open that, she will hit the land and will die.

Solution:

Step one: First you will have to make 3 trips of 1,000 apples 333 miles. Then you will be left with 2,001 apples and 667 miles to go.

Step two: Next you will have to take 2 trips of 1,000 apples 500 miles. In this manner you will be left with 1,000 apples and 167 miles to go (you have to leave an apple behind).

Step three: At last, you will have to travel the last 167 miles with one load of 1,000 apples and you will be left with 833 apples when you reach the town

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.

Solution:

99 matches.

The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 99 matches were played.

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

David to play second

Explanation:
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.

Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.

If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.

Therefore, you should suggest David to play second.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

74658

* 8 + 6 =14
* 5 - 4 = 1
* 7 = 2*4 - 1
* 4 + 6 = 10
* 7+4+6+5+8 = 30

Solution:

If the man killed himself, he would not have been able to rewind the cassette. Thus it is clear someone else killed him.

Solution:

I spoke with Joe.

Explanation:
There can be only two cases. Either Zoe is a liar or Joe is a liar.

First Case
Let us assume that Zoe is a liar and Joe is a truthful person.
If I asked the question from Joe, the answer will be yes. If I asked the question from Zoe, the answer will be no. Thus in this case, I must have asked from Joe.

Second Case
Let us assume that Zoe is a truthful and Joe is a liar.
If I asked the question from Joe, the answer will be yes. If I asked the question from Zoe, the answer will be no. Thus in this case as well, I must have asked from Joe only.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.