WHAT CAN BE A WORD A NUMBER A PE RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

The Queen And The Duke

Hint:
The Queen said that: "Until you see me again you shall not reveal to anyone what happen to me." The face of the Queen is on the money.
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A Piece On A Chessboard

Hint:
The queen
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Something That Floats Riddle

Hint:
A balloon
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A Type Of Fruit Riddle

Hint:
An orange
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Add Some Butter

Hint:
Ear of corn
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Great As Chips And Fries

Hint:
Potatoes
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Sweet And Bakes Riddle

Hint:
Potatoes
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Age Of Three Daughters Riddles

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.
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This Countrys Capital Is Canberra

Hint:
Australia
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Visit This Very Large Country

Hint:
Australia
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I Never Was But Always Will Be Riddle

Hint:
Tomorrow
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Trapdoor, Water, Wolf

Hint:
They're all spiders
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7 Guys 6 Rooms Riddle

Hint:
He never did anything with the 7th guy.
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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.
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Three Gods Riddle

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.
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