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:

charles put the ring into the box, secure it with one of your locks, and send the box to the girl firend.
She should then attach one of his own locks and return it. When you receive it again, remove your lock and send it back with her lock. Now she can unlock his own lock and retrieve the object.

Solution:

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

Solution:

He askes the god to beat him half to death.

Solution:

The password is U472BMT.

Read aloud the line and you will sound something like "U Four Seven 2 B M T".

Solution:

The code word is TIE.

Let us assume that the word is MOD. Now if you are told one character of this word, you would not be able to identify the number of vowels in the word. For example if you are told M, then the word can be MOD or AIM but they both have different number of vowels. If you are told about O, then there are two words MOD and OAT which again have different number of vowels. If you are told about D, there are two words again MOD and DUE which have different numbers of vowels again.

In such manner, all the words that have M, O or D in them can be excluded. Then only word that remains is TIE.
If you are told about T, there are two words OAT and TIE which both have two vowels.
If you are told about I, there are two words AIM and TIE which both have two vowels.
If you are told about E, there are two words DUE and TIE which both have two vowels.

Thus TIE is the code word.

Solution:

If you ask the guard who always tells the truth, he would tell you the other guard would point you to the door of death. If you ask the guard who always lies, he would tell you the opposite door of the truth-telling guard and point you to the door of death. In either case, both guards will point to the door of death so you should choose the other one.

Solution:

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combination so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a milliliter from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combination where all but one prisoner must sip from the wine. By avoiding these two types of combination you can ensure no more than 8 prisoners die.

One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.

Solution:

he was genius.

Officials must have thought he was jumping with a hostage, they would never risk giving him a faulty parachute.

Solution:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

Solution:

He simply suggested that one of the brothers will be allowed to divide the land but he must take into consideration that it will be the other one who will have a choice of selecting his land between the two halves first.

Thus both will be happy in such manner.