A Real Gun With Real Bullets
A man walks into a his bathroom and shoots himself right between the eyes using a real gun with real bullets. He walks out alive, with no blood anywhere. And no, he didn't miss and he wasn't Superman or any other caped crusader.
How did he do this?
How did he do this?
Hint:
Growing In A Field Riddle
Hint:
Zombie Date Riddle
Hint:
A Glass Of Water
A man walks into a bar and asks the bartender for a glass of water. The bartender reaches under the bar and brings out a gun and aims it at the man. The man says thank you and leaves. What happened?
Hint: Why would the man want a glass of water?
The man had the hiccups and wanted a glass of water to help get rid of them. The bartender could hear the hiccups when the man spoke, so he brought the gun out to scare the hiccups away. It worked and the man thanked him and left, no longer needing the glass of water. Did you answer this riddle correctly?
YES NO
YES NO
Can You Solve The Horse Sale Brain Teaser?
How much money did he make or lose?
Did he break even?
Hint: He didn't break even.
Age Of Three Daughters Riddles
I was visiting a friend one evening and remembered that he had three daughters. I asked him how old they were. The product of their ages is 72, he answered. Quizzically, I asked, Is there anything else you can tell me? Yes, he replied, the sum of their ages is equal to the number of my house. I stepped outside to see what the house number was. Upon returning inside, I said to my host, Im sorry, but I still cant figure out their ages. He responded apologetically, Im sorry, I forgot to mention that my oldest daughter likes strawberry shortcake. With this information, I was able to determine all three of their ages. How old is each daughter?
Hint:
3, 3, and 8. The only groups of 3 factors of 72 to have non-unique sums are 2 6 6 and 3 3 8 (with a sum of 14). The rest have unique sums:
2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13
The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer. Did you answer this riddle correctly?
YES NO
2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13
The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer. Did you answer this riddle correctly?
YES NO
David Is 6 Years Old Riddle
Hint:
David: 6
Lauren: 3
David is 3 years older than Lauren.
So when David is 100 Lauren is 97.
Answer: 97 Did you answer this riddle correctly?
YES NO
Lauren: 3
David is 3 years older than Lauren.
So when David is 100 Lauren is 97.
Answer: 97 Did you answer this riddle correctly?
YES NO
7 Guys 6 Rooms Riddle
7 Guys 6 Rooms. All men Want a Room all by Themselves.
The Hotel Manager put the first two guys in room number 1.
The Third Guy in Room Number 2.
The Fourth Guy in Room Number 3.
The Fifth Guy in Room Number 4.
The Sixth Guy in Room Number 5
But The Room Number Sixth is still Empty.
The Hotel Manager put the first two guys in room number 1.
The Third Guy in Room Number 2.
The Fourth Guy in Room Number 3.
The Fifth Guy in Room Number 4.
The Sixth Guy in Room Number 5
But The Room Number Sixth is still Empty.
Hint:
Banana Clock Riddle
Hint: 1. Look closely at the clock.
2. Number of Bananas.
3. Some thing regarding the sides of the Shapes figure.
38
Logic :
From 1st the hexagon shape has value 15.
The shape has 15 edges(6 of hexagon,5 of Pentagon and 4 of Square)
From 2nd we get value of one bunch of bananas is 4.
So each banana has value of 1.
From 3rd we get that each clock has value of 3.
Which resembles the time on the clock which is 3.
Hence by using these insights, we get the last required values as
Clock = 2 (2 in the clock)
Bananas = 3 (3 bananas in the bunch)
Hexagon = 11 (Hexagon[6 sides] and pentagon[5 sides], so 6+5=11)
So required value is
2+3+3x11=?
2+3+33=? (Multiply first - Bodmas rule)
5+33=38 and hence the answer is 38. Did you answer this riddle correctly?
YES NO
Logic :
From 1st the hexagon shape has value 15.
The shape has 15 edges(6 of hexagon,5 of Pentagon and 4 of Square)
From 2nd we get value of one bunch of bananas is 4.
So each banana has value of 1.
From 3rd we get that each clock has value of 3.
Which resembles the time on the clock which is 3.
Hence by using these insights, we get the last required values as
Clock = 2 (2 in the clock)
Bananas = 3 (3 bananas in the bunch)
Hexagon = 11 (Hexagon[6 sides] and pentagon[5 sides], so 6+5=11)
So required value is
2+3+3x11=?
2+3+33=? (Multiply first - Bodmas rule)
5+33=38 and hence the answer is 38. Did you answer this riddle correctly?
YES NO
Three Gods Riddle
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
What three questions can you ask?
What three questions can you ask?
Hint:
A possible solution is:
Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination. Did you answer this riddle correctly?
YES NO
Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination. Did you answer this riddle correctly?
YES NO
7B91011 Riddle
Mrs. Barbara Whitcombe, Ray's wife
Mr. Jason McCubbins, Ray's business partner
Mr. Harold Nichols, Ray's best friend
All three visited Mr. Whitcombe the day of his murder, but all three provided the police with stories of explanation as to the reason for their visit.
Police found Mr. Whitcombe with his wrist watch still on his right arm, a torn up picture of his wife laying on the floor beside the trash can, and an ink pen in his right hand. On the desk, the police found a name plate, a telephone that was off the hook, and a personal calendar turned to the July 5th page with 7B91011 written on it. After examining this evidence, the police knew their suspect.
Who was it?
Hint:
Jason McCubbins, Ray's business partner.
The calendar was the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must have been left-handed. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
The calendar was the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must have been left-handed. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
A Train Leaves From Halifax Riddle
Hint:
Both trains would be at the same spot when they meet therefore they are both equally close to Halifax. Did you answer this riddle correctly?
YES NO
YES NO
I Drink Milk And Eat Fish Riddle
Hint:
Five Rows Of Four Christmas Trees Riddle
Hint:
Just imagine a 5 pointed star, and then plant one tree at each point, and one tree where the sides intersect.
There are actually several distinct solutions. All of them can be constructed as follows:
Draw a nice long straight line.
Draw a second straight line that intersects the first.
Draw three more straight lines making sure each line intersects all the lines youve already drawn and avoiding any of the previous points of intersection. That is, no three lines should intersect at the same point.
With the first four lines, theres only one topologically distinct configuration, but by varying the position of the fifth line, several different distinct configurations can be created. Did you answer this riddle correctly?
YES NO
There are actually several distinct solutions. All of them can be constructed as follows:
Draw a nice long straight line.
Draw a second straight line that intersects the first.
Draw three more straight lines making sure each line intersects all the lines youve already drawn and avoiding any of the previous points of intersection. That is, no three lines should intersect at the same point.
With the first four lines, theres only one topologically distinct configuration, but by varying the position of the fifth line, several different distinct configurations can be created. Did you answer this riddle correctly?
YES NO
A Cruise Between Mexico And The USA Riddle
Hint:
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