Solution:

10,15,9,20,12

Explanation:
The score obtained on the first throw is 10.
The score obtained on the second throw is 15.
The score obtained on the third throw is 9.
The score obtained on the fourth throw is 20.
The score obtained on the fifth throw is 12.

Solution:

The wise man told them to switch the horses and then race. In this manner, both of them will be willing to race as fast as they can to prove that the other horse (his own horse) is slow.

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.

Solution:

(0! + 0! + 0!)! = 6
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3 x 3 - 3 = 6
√4 + √4 + √4 = 6
5 + 5/5 = 6
6 + 6 - 6 = 6
7 - 7/7 = 6
8 - √√(8 + 8) = 6
√(9 x 9) - √9 = 6

Solution:

The young lady was presented with a bill which belonged to the elderly woman. She had asked the waiter to collect the bill from her daughter.

Solution:

Place 1 white marble in one bowl, and place the rest of the marbles in the other bowl (49 whites, and 50 blacks).

This way you begin with a 50/50 chance of choosing the bowl with just one white marble, therefore life! BUT even if you choose the other bowl, you still have ALMOST a 50/50 chance at picking one of the 49 white marbles.

Solution:

2519

The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.

Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.

Solution:

From left to right
#1 Mr. Reed Richard – Blue Marbles
#2 Ms. Sue Storm – Red Marbles
#3 Mr. Johnny Storm – White Marbles
#4 Mr. Ben Grimm – Green Marbles.

Let us label the houses from left to right as #1, #2, #3, #4. Now let us split the hints in different category.
a) Mr. Reed Richard’s house is to the left of the green marbles one
b) The third one is made of white marbles
c) Ms. Sue Storm owns a red marbles house
d) Mr. Johnny Storm does not live on either ends
e) Mr. Johnny Storm lives right of the blue marbles house
f) Mr. Ben Grimm lives in the fourth house
g) The first house is not made up of red marbles

By seeking all the (g) and (b), we are sure that the #1 is not made up of red marbles and nor is the #3. Now with (f) we know that Mr. Ben Grimm lives in #4 made up of green marbles and keeping in mind (c) #2 must be made of Red Marbles and Ms. Sue Storm lives there. Now moving to the hint (d) Mr. Johnny Storm must be living in #3 and adding up with hint (b), it must be made from white marbles. Now by hint (a) #4 must be made up of green marbles (or Mr. Reed Richard’s house can’t be on its left). Also by (f) Mr. Ben Grimm lives there. All of it leaves Mr. Reed Richards living in #1 with blue marbles construction.

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:

The poison was not in pills but in water.