I Am A Solitary Word 5 Letters Long B Riddles To Solve
Solving I Am A Solitary Word 5 Letters Long B Riddles
Here we've provide a compiled a list of the best i am a solitary word 5 letters long b puzzles and riddles to solve we could find.Our team works hard to help you piece fun ideas together to develop riddles based on different topics. Whether it's a class activity for school, event, scavenger hunt, puzzle assignment, your personal project or just fun in general our database serve as a tool to help you get started.
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A Solitary Word 5 Letters Long Riddle
I am a solitary word, 5 letters long.
Behead me, and I am still the same.
Behead me again, and I am still the same.
What am I?
Behead me, and I am still the same.
Behead me again, and I am still the same.
What am I?
Hint: Reread the first line.
Alone.
Behead me and I am Lone.
Behead me again and I am One. Did you answer this riddle correctly?
YES NO
Behead me and I am Lone.
Behead me again and I am One. Did you answer this riddle correctly?
YES NO
Words And Letters Riddle
Hint:
Longest Word In The Dictionary
Hint:
The Longest, Non-scientific Word
Hint:
What is the longest, non-scientific word to have all of its letters in alphabetical order? Did you answer this riddle correctly?
YES NO
YES NO
Longest Word In English Riddle
Hint:
What Is A Word Made Up Of 4 Letters Riddle
Hint:
The word 'what' has 4 letters in it, 'yet' has three, 'sometimes' has 9, 'then' has 4, 'rarely' has 6, and 'never' has 5. Did you answer this riddle correctly?
YES NO
YES NO
A Word I Know Riddle
Hint:
Remove My Letters Riddle
I am a five letter word. If you remove all my letters except the first, I sound the same! What am I?
Hint:
Twenty Six Letters Riddle
Hint:
Bus Driver Eyes
Hint:
The eye color of the reader of this problem. The first sentence is the key: "You are the bus driver". Did you answer this riddle correctly?
YES NO
YES NO
A Word Referring To Women Riddle
Hint:
In A Basket All Alone Riddle
Once a little babe was drifting
In a basket all alone,
When the king's fair daughter found him-
Wished to have him for her own.
So she found the baby's mother
For his nurse that very day
But when he had grown to manhood,
He his people led away.
Who was he?
In a basket all alone,
When the king's fair daughter found him-
Wished to have him for her own.
So she found the baby's mother
For his nurse that very day
But when he had grown to manhood,
He his people led away.
Who was he?
Hint:
Burns Me There Riddle
Hint:
Rearrange the letters of BURN ME THERE and they spell out the words THREE NUMBERS! Did you answer this riddle correctly?
YES NO
YES NO
10 Boxes Riddle
There are ten boxes containing some balls. Each of the ball weighs exactly 10 grams. One of those boxes have defective balls (all the defective balls weigh 9 grams each).
An electronic weighing machine is provided to you and you are allowed only one chance of weighing on it.
How will you find out which box has defective balls ?
An electronic weighing machine is provided to you and you are allowed only one chance of weighing on it.
How will you find out which box has defective balls ?
Hint:
Let us simplify boxes by naming them from 1 to 10.
Now the trick here is to pick different number of balls from different boxes. So to simplify things, we will pick balls corresponding to box number.
Thus, pick 1 ball from Box 1, 2 balls from box 2, 3 balls from box 3 and so on. You will have 55 balls altogether. Now, put them all in the balance.
If all balls were weighing accurate 10 grams, the total weight of the 55 balls would have been 550 grams. But one of the box must have had the defective balls.
Suppose if the defective balls were in box number 2, then the total weight will be 2 grams less than 550. If the defective balls were in box 8, the total weight will be less than 8 grams from 550. In this way, you will be able to identify which box has the defective balls. Did you answer this riddle correctly?
YES NO
Now the trick here is to pick different number of balls from different boxes. So to simplify things, we will pick balls corresponding to box number.
Thus, pick 1 ball from Box 1, 2 balls from box 2, 3 balls from box 3 and so on. You will have 55 balls altogether. Now, put them all in the balance.
If all balls were weighing accurate 10 grams, the total weight of the 55 balls would have been 550 grams. But one of the box must have had the defective balls.
Suppose if the defective balls were in box number 2, then the total weight will be 2 grams less than 550. If the defective balls were in box 8, the total weight will be less than 8 grams from 550. In this way, you will be able to identify which box has the defective balls. Did you answer this riddle correctly?
YES NO
A Rickety Bridge Riddle
Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?
Hint:
17 mins.
The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.
Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins Did you answer this riddle correctly?
YES NO
The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.
Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
Total time = 2 + 2 + 10 + 1 + 2 = 17 mins Did you answer this riddle correctly?
YES NO
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