Solution:

John Walters was drowned in the quicksand.

Solution:

20
Just count the nines in the numbers: 6, 16, 26, 36, 46, 56, 76, 86, 96, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.
Note: 66 contains TWO nines

Solution:

Sansa has the siamese (conjoined) twin sister.

Solution:

The attached picture gives away the solution. If you can find another solution, do send us yours.

Solution:

8/(1−1/5)

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:

Upon reaching each room, he found out that the window facing the dead body was closed. He had to open the window himself. This tells that there was somebody else with him who closed the window. If he would have committed suicide, at least one window would have been open.

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:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

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:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.