138 Reasoning question on puzzles

Here are some reasoning questions on puzzles for you to practice and challenge your mind:

Example 1: The Age Puzzle

Question:
There are three people in a room: Alex, Ben, and Carl. The sum of Alex and Ben’s ages is 50. Ben is 10 years younger than Carl, and Carl is 20 years older than Alex. What are their ages?

Solution:
Let Alex’s age be x.
Ben’s age will be 50 – x (since Alex and Ben’s ages sum to 50).
We know Ben is 10 years younger than Carl, so Carl’s age will be 50 – x + 10.
We also know Carl is 20 years older than Alex, so:
50 – x + 10 = x + 20.
Solving for x:
50 – x + 10 = x + 20
60 – x = x + 20
60 – 20 = 2x
40 = 2x
x = 20.
So, Alex is 20, Ben is 30 (50 – 20), and Carl is 40 (20 + 20).

Example 2: The Bridge Crossing Puzzle

Question:
Four people need to cross a bridge at night. There is a single flashlight, and the bridge is too dangerous to cross without it. The group has four people with different crossing times:

• Person A takes 1 minute to cross.
• Person B takes 2 minutes to cross.
• Person C takes 5 minutes to cross.
• Person D takes 10 minutes to cross.
  Only two people can cross at a time. What is the fastest way for all four to cross the bridge?

Solution:

1. A and B cross together (2 minutes).
2. A returns with the flashlight (1 minute).
   Total time: 3 minutes.
3. C and D cross together (10 minutes).
   Total time: 13 minutes.
4. B returns with the flashlight (2 minutes).
   Total time: 15 minutes.
5. A and B cross together again (2 minutes).
   Total time: 17 minutes.

Thus, the fastest time for all four to cross the bridge is 17 minutes.

Example 3: The Riddle of the Hats

Question:
There are three people, A, B, and C, standing in a line, each wearing a hat. They are told that there are two white hats and one black hat, and they must figure out which hat they are wearing. The people can see the hats of the others, but not their own. After a while, they all know what color hat they are wearing. What is the reasoning behind this?

Solution:
A can see the hats of B and C. If both B and C are wearing white hats, A would know they must be wearing the black hat. However, A doesn’t immediately declare the color of their hat. This means that either B or C is wearing a black hat.
After seeing A’s hesitation, B and C can deduce that one of them must be wearing the black hat. Eventually, B and C figure out that they are wearing the white hats, and A realizes they must be wearing the black hat.

Example 4: The Coin Puzzle

Question:
You have 12 coins, and one of them is either heavier or lighter than the others. You can only use a balance scale three times. How can you identify the odd coin and determine whether it is heavier or lighter?

Solution:

1. First weighing: Divide the 12 coins into three groups of 4. Weigh two groups against each other.
   • If they balance, the odd coin is in the group that was not weighed.
   • If they don’t balance, the odd coin is in the heavier or lighter group, depending on the result.
2. Second weighing: Take the group that contains the odd coin and divide it into three groups of 1, 1, and 2 coins. Weigh the 1-coin groups against each other.
   • If they balance, the odd coin is in the group of 2 coins.
   • If they don’t balance, the odd coin is in the group that was heavier or lighter, depending on the result.
3. Third weighing: Take the two coins from the group of 2 and weigh them against each other.
   • If they balance, the odd coin is the one you didn’t weigh.
   • If they don’t balance, the odd coin is the heavier or lighter one.

In just three weighings, you can identify the odd coin and whether it is heavier or lighter.

Example 5: The Family Puzzle

Question:
A man is looking at a picture of someone. His friend asks, “Who are you looking at?” The man replies, “Brothers and sisters, I have none. But the father of the person in the picture is my father’s son.” Who is the person in the picture?

Solution:
The person in the picture must be the man’s son.
Here’s the reasoning:

• The man has no brothers, so “my father’s son” must refer to himself.
• Therefore, the father of the person in the picture (the man’s son) is “my father’s son,” which means the person in the picture is his son.

Example 6: The Water Jug Puzzle

Question:
You have two jugs: one holds 5 liters, and the other holds 3 liters. You need to measure exactly 4 liters of water. How can you do it?

Solution:

1. Fill the 5-liter jug completely.
2. Pour the water from the 5-liter jug into the 3-liter jug until the 3-liter jug is full.
3. Now you have 2 liters in the 5-liter jug (5 – 3 = 2).
4. Empty the 3-liter jug.
5. Pour the remaining 2 liters from the 5-liter jug into the 3-liter jug.
6. Fill the 5-liter jug again.
7. Now pour the water from the 5-liter jug into the 3-liter jug until it is full. The 3-liter jug already contains 2 liters, so only 1 more liter can be added to it.
8. Now, you have exactly 4 liters of water in the 5-liter jug.

Example 7: The Light Switch Puzzle

Question:
You are in a room with three light switches. One of them controls a light bulb in another room. You cannot see the light bulb from the room with the switches. How can you determine which switch controls the light bulb if you can only enter the other room once?

Solution:

1. Turn on the first switch and leave it on for a few minutes.
2. Turn off the first switch and turn on the second switch.
3. Immediately go to the other room:
   • If the light is on, the second switch controls the light bulb.
   • If the light is off but warm, the first switch controls the light bulb (because it was on long enough to heat the bulb).
   • If the light is off and cold, the third switch controls the light bulb.

Example 8: The Odd Man Out Puzzle

Question:
You have five people: A, B, C, D, and E. They are sitting in a circle. A is sitting next to B, but not next to C. C is sitting next to E, but not next to D. D is sitting next to A, but not next to B. Who is sitting next to C?

Solution:
From the clues:

• A is next to B but not C, so A and B are next to each other but not next to C.
• C is next to E but not D, so C and E must be next to each other.
• D is next to A but not B, so D and A are next to each other, but D is not next to B.

Arranging these positions, you’ll find:

• A sits next to B and D.
• B sits next to A.
• C sits next to E.
• D sits next to A.
• E sits next to C.

So, E is sitting next to C.

Example 9: The Room of 3 Doors Puzzle

Question:
You are in a room with 3 doors. Behind one door is a treasure, behind another door is a lion, and behind the third door is an empty room. You don’t know which door leads to what. You can ask only one question to one of the guards who each stand in front of a door. What question can you ask to find the door that leads to the treasure?

Solution:
You can ask any guard:
“If I were to ask the guard in front of the other door which door leads to the treasure, which door would they point to?”

• If you ask the guard in front of the treasure door, they will point to the empty room (because the other guard would point there).
• If you ask the guard in front of the empty room, they will point to the lion (because the other guard would point to the treasure).
• In either case, the guard you ask will point to the wrong door, so you just choose the other one.

Example 10: The River Crossing Puzzle

Question:
You have a wolf, a goat, and a cabbage. You need to cross a river with these three items, but you can only take one at a time. If you leave the wolf with the goat, the wolf will eat the goat. If you leave the goat with the cabbage, the goat will eat the cabbage. How do you get all three across the river safely?

Solution:

1. First, take the goat across the river and leave it on the other side.
2. Go back across the river alone.
3. Then take the wolf across the river.
4. Bring the goat back with you to the original side.
5. Take the cabbage across the river and leave it with the wolf.
6. Finally, return across the river alone and bring the goat across.

All three (the wolf, goat, and cabbage) are safely across.

Example 11: The Coin Flip Puzzle

Question:
You have two identical coins, but one of them is biased and always lands heads, and the other is fair. You do not know which coin is which. How can you determine which coin is biased with just one flip of one coin?

Solution:
You can flip both coins once.

• If both coins land the same (either both heads or both tails), the biased coin is the one that landed heads (because the biased coin always lands heads).
• If one coin lands heads and the other lands tails, then the biased coin is the one that landed heads.

Example 12: The Three Sons Puzzle

Question:
A man has three sons. The first son says, “I am the oldest.” The second son says, “I am the youngest.” The third son says, “I am the middle.” The man knows that only one of them is telling the truth. Which son is which?

Solution:

• If the first son is telling the truth (he is the oldest), then the second son (who says he is the youngest) must be lying, and the third son must be the middle one.
• If the second son is telling the truth (he is the youngest), then the first son is the oldest, and the third son is the middle one. But if the first son is the oldest, then the third son cannot be the middle because the third son is lying.
• The only possibility left is that the third son is telling the truth. Therefore, the third son is the middle, the first son is the oldest, and the second son is the youngest.

Example 13: The Prisoners and Hats Puzzle

Question:
Five prisoners are each given a hat that is either red or blue. They are lined up so they can see the hats of the people in front of them, but not their own hats. The warden tells them that at least one prisoner is wearing a red hat. The prisoners can only say one word: “Red” if they think they are wearing a red hat, and “Blue” if they think they are wearing a blue hat. They are not allowed to talk to each other. If any prisoner says the wrong color, all the prisoners are executed. What strategy should the prisoners use to ensure their survival?

Solution:
The strategy is for the prisoners to count the number of red hats they see in front of them.

• The last prisoner (who can see all the hats) will say “Red” if they see an odd number of red hats and “Blue” if they see an even number.
• Each subsequent prisoner counts how many red hats they see in front of them, and if the number of red hats is odd (and the last prisoner said “Red”) or even (and the last prisoner said “Blue”), they deduce their own hat’s color.
• This strategy ensures that every prisoner survives.

Example 14: The Family Relationship Puzzle

Question:
In a family, there are two fathers, two sons, and one grandfather, but only four people. How is this possible?

Solution:
The family consists of a grandfather, his son (who is both a father and a son), and his grandson. So:

• The grandfather is one father.
• The father is the second father and the son of the grandfather.
• The grandson is the second son.

Example 15: The Mismatched Socks Puzzle

Question:
You are in a dark room with 100 socks: 50 red socks and 50 blue socks. You must pick two socks at random, but you can’t see the color of the socks. What is the minimum number of socks you need to pick to ensure that you have at least one matching pair?

Solution:
You need to pick 3 socks.
Here’s why:

• If the first two socks are different colors, the third one will match one of them.
• So, by picking 3 socks, you are guaranteed to have at least one matching pair.

Example 16: The Light Bulb Puzzle

Question:
You have three light bulbs in a room and three switches outside the room. Each switch controls one bulb, but you don’t know which switch controls which bulb. You can turn the switches on and off as many times as you want, but you can only enter the room once. How do you determine which switch controls which bulb?

Solution:

1. Turn on the first switch and leave it on for a few minutes.
2. Turn off the first switch and immediately turn on the second switch.
3. Go into the room:
   • The bulb that is on is controlled by the second switch.
   • The bulb that is off but warm is controlled by the first switch (because it was on long enough to heat up).
   • The bulb that is off and cold is controlled by the third switch.

Example 17: The Family Riddle

Question:
A man is looking at a picture of someone. His friend asks, “Who are you looking at?” The man replies, “Brothers and sisters, I have none. But the father of the person in the picture is my father’s son.” Who is the person in the picture?

Solution:
The person in the picture must be the man’s son.
Here’s the reasoning:

• “My father’s son” refers to the man himself (since he has no brothers).
• Therefore, the person in the picture is his son.

Example 18: The Water Jug Puzzle (Revised)

Question:
You have two jugs: one holds 7 liters and the other holds 5 liters. You need to measure exactly 3 liters of water. How do you do it?

Solution:

1. Fill the 7-liter jug completely.
2. Pour the water from the 7-liter jug into the 5-liter jug until the 5-liter jug is full.
   Now you have 2 liters left in the 7-liter jug (7 – 5 = 2).
3. Empty the 5-liter jug.
4. Pour the 2 liters from the 7-liter jug into the 5-liter jug.
5. Fill the 7-liter jug again.
6. Pour water from the 7-liter jug into the 5-liter jug, which already contains 2 liters, until it is full.
   Now the 7-liter jug has exactly 3 liters of water remaining.

Example 19: The Two Doors Puzzle

Question:
You are in a room with two doors. One door leads to freedom, and the other leads to certain death. There are two guards: one guard always tells the truth, and the other always lies. You don’t know which guard is which. You can ask one question to one of the guards to determine which door leads to freedom. What question do you ask?

Solution:
You can ask either guard, “If I were to ask the other guard which door leads to freedom, what would they say?”

• If you ask the truthful guard, they will tell you the door that leads to death (because the liar would point to that door).
• If you ask the liar, they will lie about what the truthful guard would say and also point to the door that leads to death.
  So, in either case, the answer you receive will be the door that leads to death. Therefore, you should choose the opposite door.

Example 20: The Coin Puzzle (Revised)

Question:
You have 8 coins. One of them is fake and weighs less than the others. You have a balance scale, and you can only use it twice. How do you find the fake coin?

Solution:

1. Divide the 8 coins into three groups: two groups with 3 coins each and one group with 2 coins.
2. Weigh the two groups of 3 coins against each other.
   • If they balance, the fake coin is in the group of 2 coins.
   • If they don’t balance, the fake coin is in the lighter group of 3 coins.
3. Take the group with the fake coin (either the 3-coin group or the 2-coin group) and weigh 2 coins against each other.
   • If they balance, the remaining coin is the fake one.
   • If they don’t balance, the lighter coin is the fake one.

Example 21: The Missing Dollar Puzzle

Question:
Three friends go to a hotel room and pay $30 for the room. Later, the hotel manager realizes there was a mistake, and the room cost only $25. The manager gives $5 to the bellboy and asks him to return it to the three friends. The bellboy, thinking it’s hard to divide $5 among three people, keeps $2 for himself and gives $1 back to each of the three friends. Now, each friend has paid $9 (totaling $27), and the bellboy has $2, which makes $29. Where is the missing dollar?

Solution:
There is no missing dollar. The trick is in the way the problem is framed.

• The three friends paid a total of $27. Out of that $27, $25 went to the hotel for the room, and the remaining $2 went to the bellboy.
• The mistake comes from adding the $2 that the bellboy kept to the $27. Instead, the $27 already includes the $2 that the bellboy kept. So, there’s no missing dollar.

Example 22: The Paradox of the Prisoner’s Hat

Question:
A prisoner is told that he will be executed, but there’s a twist: there are two hats, one red and one blue, and the prisoner has to choose one. If he guesses correctly, he will be set free; if he guesses incorrectly, he will be executed. The prisoner is not told which color hat he will wear, but he can see the hat of the person next to him. How can the prisoner guarantee his survival?

Solution:
This is a paradox that depends on what the prisoner chooses to do in advance. The prisoner can use a predetermined strategy:

• Before the hats are placed, the prisoner decides that he will always guess the opposite color of the hat he sees on the other person’s head.
• The idea is that, by guessing the opposite color, the prisoner creates a 50% chance for survival based on a system of logic and assumptions about the other person’s guess, as it’s essentially a paradox of deductive reasoning.

Example 23: The Family Relationship Puzzle (Revisited)

Question:
A man is looking at a picture of someone. His friend asks, “Who are you looking at?” The man replies, “Brothers and sisters, I have none. But the father of the person in the picture is my father’s son.” Who is the person in the picture?

Solution:
The person in the picture must be the man’s son.

• “My father’s son” refers to the man himself because he has no brothers.
• Therefore, the person in the picture is his son.

Example 24: The 5 Litre Jug Puzzle

Question:
You have a 5-litre jug and a 3-litre jug. Neither jug has any markings on it. How can you measure exactly 4 litres of water using just these two jugs?

Solution:

1. Fill the 5-litre jug completely (now you have 5 litres).
2. Pour the water from the 5-litre jug into the 3-litre jug until the 3-litre jug is full.
3. Now you have 2 litres of water left in the 5-litre jug (5 – 3 = 2).
4. Empty the 3-litre jug.
5. Pour the remaining 2 litres from the 5-litre jug into the 3-litre jug.
6. Fill the 5-litre jug again.
7. Pour the water from the 5-litre jug into the 3-litre jug, which already contains 2 litres. Now, the 3-litre jug is full, and exactly 4 litres will remain in the 5-litre jug.

Example 25: The Two Sisters Puzzle

Question:
You are in a room with two sisters. One of them always tells the truth, and the other always lies. You need to find out which sister is which, but you can only ask them one question. What question should you ask?

Solution:
You can ask either sister:
“If I were to ask your sister which one of you always tells the truth, what would she say?”

• If you ask the truthful sister, she will tell you that the other sister is the liar (because the liar would lie about who tells the truth).
• If you ask the lying sister, she will lie about what the truthful sister would say, and point to herself as the truthful one.
  In both cases, you can deduce that the sister they point to is the liar.

Example 26: The Light Bulb and Switch Puzzle (Revisited)

Question:
You have 3 switches outside a room, each controlling one of 3 light bulbs inside the room. You can only enter the room once, and you don’t know which switch controls which bulb. How do you figure out which switch controls which bulb?

Solution:

1. Turn on the first switch and leave it on for a few minutes to allow the bulb to warm up.
2. Turn off the first switch and turn on the second switch.
3. Quickly go into the room:
   • The bulb that is on is controlled by the second switch (since it was turned on last).
   • The bulb that is off but warm is controlled by the first switch (since it was on earlier).
   • The bulb that is off and cold is controlled by the third switch (because it was never turned on).

Example 27: The Three Hats Puzzle

Question:
Three people, A, B, and C, are each wearing a hat. There are two red hats and one blue hat. They are all told they cannot see their own hats but can see the hats of others. They are then asked, one by one, if they know the color of their own hat. The responses are:

• A says, “I don’t know.”
• B says, “I don’t know.”
• C says, “I know.”

What color is C’s hat?

Solution:

• A sees both B and C’s hats. If they were both wearing blue hats, A would know their hat must be red, but A says, “I don’t know.” So, A sees at least one red hat.
• B then knows that A must see at least one red hat. If B were wearing a blue hat, B would know their own hat must be red (because A would have seen both blue hats). Since B also says, “I don’t know,” B must be wearing a red hat.
• Since B and A are both wearing red hats, C must be wearing the blue hat.

So, C is wearing the blue hat.

Example 28: The 3 Switches Puzzle (Alternate Version)

Question:
You are in a room with 3 switches, each controlling one of 3 light bulbs. You can flip the switches and enter the room only once. How can you determine which switch controls which bulb?

Solution:

1. Flip the first switch and leave it on for a while.
2. After a minute, flip the first switch off and turn the second switch on.
3. Enter the room:
   • The bulb that is on is controlled by the second switch (because it’s the only one that’s on).
   • The bulb that is off but warm is controlled by the first switch (because it was on long enough to heat up).
   • The bulb that is off and cold is controlled by the third switch (because it was never turned on).

Example 29: The Family Relation Puzzle

Question:
A woman gives birth to a son, and the son grows up and gets married. The woman then gives birth to a daughter-in-law. How is the woman related to her daughter-in-law?

Solution:
The woman is the mother-in-law of her daughter-in-law.

Example 30: The Three Sons and Two Statements Puzzle

Question:
A man has three sons. The first son says, “I am the oldest.” The second son says, “I am the youngest.” The third son says, “I am the middle.” The man knows that only one of the sons is telling the truth. Which son is which?

Solution:

• If the first son is telling the truth, he must be the oldest.
• The second son claims to be the youngest, which means the third son must be the middle.
• Since only one son is telling the truth, the third son must be lying, meaning the first son is the oldest and the second son is the youngest.
• So, the first son is the oldest, the second son is the youngest, and the third son is the middle.

Example 31: The Coin Puzzle (Revisited)

Question:
You have 10 coins placed in a line. Nine of the coins are heads up, and one is tails up. Your task is to flip exactly 5 coins to make all the coins show heads up. How do you do it?

Solution:

1. Flip any 5 coins.
2. No matter which 5 coins you flip, you will have 4 heads flipped and 1 tail flipped.
3. Now, you just need to flip the one coin that is tails up.
   This will give you 5 heads.

Example 32: The Five People Puzzle

Question:
Five people are standing in a line. A man, B woman, and C child, in that order. A is the tallest, B is the second tallest, and C is the shortest. C says, “I am taller than A,” and B says, “I am taller than C.” Who is the shortest and who is the tallest?

Solution:

• C is the shortest, since they are clearly the shortest by being placed at the end.
• A is the tallest, since A is standing at the front.
• The statements made by B and C are just part of the puzzle narrative, not factually true. It’s just part of their respective parts.

Example 33: The Farmer’s Problem

Question:
A farmer has a chicken, a fox, and a bag of corn. He needs to cross a river but can only carry one item at a time. If he leaves the chicken with the fox, the fox will eat the chicken. If he leaves the chicken with the corn, the chicken will eat the corn. How does he get all three across safely?

Solution:

1. First, take the chicken across the river and leave it on the other side.
2. Go back and take the fox across the river.
3. Bring the chicken back to the starting side.
4. Take the corn across the river and leave it with the fox.
5. Go back across the river to get the chicken, and then take it across.

Now all three items—the chicken, the fox, and the corn—are safely across the river.

Example 34: The Five Friends Puzzle

Question:
Five friends, A, B, C, D, and E, are sitting in a row. A is sitting next to B, but not next to C. C is sitting next to D. Who is sitting in the middle?

Solution:

• A and B are next to each other, so either A or B must be on either end.
• C is sitting next to D, so C and D must be together.
• This leaves A, B, C, D, and E sitting in this order: A, C, D, B, E.
• C is sitting in the middle.

Example 35: The 12 Marbles Puzzle

Question:
You have 12 marbles. One of them is heavier or lighter than the rest, but you don’t know which one. You have a balance scale, and you can use it three times. How do you find the odd marble?

Solution:

1. Divide the 12 marbles into three groups of 4 marbles each.
2. Weigh two of the groups of 4 marbles against each other.
   • If the two groups balance, the odd marble is in the group that wasn’t weighed.
   • If they don’t balance, the odd marble is in the heavier or lighter group, depending on which one is different.
3. Take the group of 4 marbles that contains the odd one and divide it into two groups of 2 marbles each.
4. Weigh the two groups of 2 marbles.
   • If they balance, the odd marble is one of the remaining 2 marbles.
   • If they don’t balance, the odd marble is in the heavier or lighter group.
5. Weigh the final two marbles to find the odd one.

Example 36: The 100 Prisoners Puzzle

Question:
100 prisoners are each given a numbered hat, either red or blue. They can see the hats of others, but not their own. They are told that if they guess the color of their own hat correctly, they will be freed. They are allowed to say their guess in any order, but if they are wrong, they are executed. What strategy can they use to guarantee the maximum number of survivors?

Solution:
The prisoners can use binary numbers. Here’s the strategy:

1. Beforehand, the prisoners agree on a rule: one prisoner will call out the parity (even or odd) of the number of red hats they see.
2. The first prisoner to guess (who can see everyone else’s hats) will count the number of red hats.
   • If they see an even number, they will say “Red” (for even), and if they see an odd number, they will say “Blue.”
3. Each subsequent prisoner can then deduce their own hat color based on the parity of the red hats they see and the first prisoner’s statement.
4. By following this method, 99 prisoners will be guaranteed to survive, and only one might not.

Example 37: The Three Switches Puzzle (Version 3)

Question:
You have three switches outside a room, each controlling one of three light bulbs inside the room. You can only enter the room once. How can you determine which switch controls which bulb?

Solution:

1. Turn on the first switch and leave it on for several minutes.
2. After the 5 minutes, turn off the first switch and turn on the second switch.
3. Now, enter the room:
   • The bulb that is on is controlled by the second switch (since you turned it on just before entering).
   • The bulb that is off but warm is controlled by the first switch (because it was on long enough to heat up).
   • The bulb that is off and cold is controlled by the third switch (since it was never turned on).

Example 38: The Stolen Goods Puzzle

Question:
A thief is caught after stealing three objects: a book, a clock, and a wallet. The police ask him, “Which one of these did you steal first?” The thief replies, “I stole the wallet second.” The police then ask, “Which one did you steal last?” The thief replies, “I stole the book first.” Can the thief be lying? If so, which object did he steal first?

Solution:
The thief’s statements cannot both be true.

• If the thief stole the wallet second, then he could not have stolen the book first. He must have stolen the book last and the clock first.
• So, the thief stole the clock first, the wallet second, and the book last.

Example 39: The Two Fathers and Two Sons Puzzle

Question:
A man and his son are sitting together. They are both fathers and both sons. How is this possible?

Solution:
The man is a grandfather. The son is his son, and the grandfather is also a father. This way, the two fathers and two sons are:

• The grandfather (father of the son),
• The father (the son of the grandfather),
• The son (the grandson).

Example 40: The Four Friends Puzzle

Question:
Four friends, A, B, C, and D, are sitting in a row. A is sitting next to B but not next to C. C is sitting next to D. Who is sitting in the middle?

Solution:

• A and B must be seated together, and C is seated next to D.
• The seating arrangement must be A, C, D, B.
• So, C is sitting in the middle.

Example 41: The Letter Puzzle

Question:
You have six letters: A, B, C, D, E, and F. How many ways can you arrange these six letters such that no two vowels (A, E) are adjacent?

Solution:

• First, arrange the consonants (B, C, D, F). There are 4! = 24 ways to arrange the consonants.
• Next, place the vowels (A, E) in the remaining 5 positions. There are 4 gaps between the consonants where the vowels can be placed.
• The vowels can be arranged in 2! = 2 ways in these gaps.

So, the total number of ways to arrange the six letters such that no two vowels are adjacent is:
24 * 2 = 48 ways.

Example 42: The Missing Dollar Puzzle (Version 2)

Question:
Three people check into a hotel room that costs $30. They each contribute $10. The hotel clerk later realizes that the room rate should have only been $25, so he gives $5 to the bellboy to return to the three guests. The bellboy, however, decides to keep $2 for himself and give $1 back to each guest. Now each guest has paid $9 (totaling $27), and the bellboy has $2. What happened to the missing dollar?

Solution:
There’s no missing dollar. The confusion comes from how the amounts are added.

• The guests paid a total of $27, which includes the $25 for the room and the $2 the bellboy kept.
• The remaining $3 was returned to the guests, so there’s no “missing” dollar. The $27 accounts for the room cost and the bellboy’s $2.

Example 43: The 8-coin Puzzle

Question:
You have 8 identical-looking coins, one of which is fake and lighter than the others. You have a balance scale, and you are allowed to use it twice. How do you find the fake coin?

Solution:

1. Divide the 8 coins into three groups: 3 coins in the first group, 3 coins in the second group, and 2 coins in the third group.
2. Weigh the first two groups (3 coins each) against each other.
   • If the groups are equal, the fake coin is in the third group (the 2 remaining coins).
   • If one group is lighter, the fake coin is in that group.
3. Now, you are left with 3 coins (either from the lighter group or the 2 unweighed coins). Weigh two of them against each other.
   • If they balance, the remaining coin is the fake one.
   • If they don’t balance, the lighter coin is the fake one.

Example 44: The Birthday Puzzle

Question:
A man was born on January 1st, 1900. He will die on his 100th birthday. When will this happen?

Solution:
The man will die on January 1st, 2000.
He was born on January 1st, 1900, and 100 years later, he will celebrate his 100th birthday on January 1st, 2000.

Example 45: The Two Sons Puzzle

Question:
A man has two sons. One is 12 years old, and the other is half his age. How old will the older son be when the younger son turns 18?

Solution:

• The older son is 12, and the younger son is half his age, so the younger son is 6 years old.
• In 12 years, the younger son will turn 18.
• Since the older son is 6 years older than the younger son, the older son will be 18 + 6 = 24 years old when the younger son turns 18.

Example 46: The Family Relationship Puzzle (Version 2)

Question:
A woman has two daughters. One daughter says, “My sister is younger than me,” and the other daughter says, “I am older than my sister.” What is the relationship between the two daughters?

Solution:
The puzzle is a trick. Both daughters are telling the truth.

• The statement “I am older than my sister” is said by the older daughter.
• The statement “My sister is younger than me” is said by the younger daughter.
• Therefore, the relationship is that the two daughters are simply siblings with one being older than the other.

Example 47: The Secret Code Puzzle

Question:
There is a secret code where numbers 1 to 5 represent letters A to E, respectively. What is the secret word for the code 1-4-2-5-3?

Solution:
Using the number-letter key:

• 1 = A
• 4 = D
• 2 = B
• 5 = E
• 3 = C

The secret word is ADBEC.

Example 48: The Detective Puzzle

Question:
A woman is found murdered. The detective interrogates the husband and staff and is given the following alibis:

• The husband said he was sleeping.
• The cook said he was cooking dinner.
• The gardener said he was planting seeds.
• The maid said she was setting the table.
• The butler said he was polishing the silverware.

The detective immediately knew who the murderer was. Who did it, and how did the detective know?

Solution:
The maid is the murderer.
She said she was setting the table, but the murder occurred before dinner, and the table wouldn’t have been set before then.

Example 49: The Digit Puzzle

Question:
I am a three-digit number. My tens digit is five more than my ones digit. My hundreds digit is eight less than my tens digit. What number am I?

Solution:
Let’s break it down:

• Let the hundreds digit be HHH, the tens digit be TTT, and the ones digit be OOO.
• From the puzzle:
  1. T=O+5T = O + 5T=O+5
  2. H=T−8H = T – 8H=T−8

Now, solve:

• Since T=O+5T = O + 5T=O+5, and H=T−8H = T – 8H=T−8, substitute TTT into the second equation:
  • H=(O+5)−8=O−3H = (O + 5) – 8 = O – 3H=(O+5)−8=O−3

Now check possible values for OOO:

• If O=4O = 4O=4, then T=4+5=9T = 4 + 5 = 9T=4+5=9 and H=9−8=1H = 9 – 8 = 1H=9−8=1.
• Therefore, the number is 194.

Example 50: The 7 Bridges Puzzle (Konigsberg Bridge Problem)

Question:
In the city of Konigsberg, there were seven bridges connecting different parts of the city. The challenge was to walk through the city in such a way that you cross each bridge exactly once. Is it possible to do this?

Solution:
The problem is a classic example of a graph theory problem. The key is to look at the degree (number of bridges) connected to each landmass. If all landmasses (vertices) have an even degree (an even number of bridges), then it’s possible to walk through all the bridges exactly once. However, in this case, the four landmasses have an odd number of bridges, so it’s impossible to walk across all the bridges exactly once without retracing your steps.

Example 51: The Prisoners and Hats Puzzle (Revisited)

Question:
100 prisoners are each given a hat, either black or white. Each prisoner can see everyone else’s hat, but not their own. They are told that they must guess the color of their own hat. If they guess correctly, they will be set free. What strategy can they use to maximize the number of prisoners who guess their hat color correctly?

Solution:
The strategy is to have the first prisoner (who is the last to guess) announce the parity (even or odd) of the number of black hats they see. This allows the remaining prisoners to deduce their own hat color by keeping track of the number of black hats they can see and matching it to the initial parity announcement. By following this strategy, 99 prisoners can be guaranteed to guess their hat color correctly, while the first prisoner may have to guess.

Example 52: The Three Children Puzzle

Question:
A man has three children. The first child says, “I am younger than my brother.” The second child says, “I am older than my sister.” The third child says, “I am not the youngest.” How old are the children?

Solution:
The third child is lying, since it says, “I am not the youngest,” which is contradictory.

• The first child says, “I am younger than my brother,” so they must be the youngest.
• The second child says, “I am older than my sister,” so they must be the oldest.
• Thus, the third child is the middle one.
  The ages, from youngest to oldest, are: youngest, middle, oldest.

Example 53: The Family Tree Puzzle

Question:
A man is looking at a picture. His friend asks, “Who are you looking at?” The man replies, “Brothers and sisters, I have none. But the father of the person in the picture is my father’s son.” Who is the person in the picture?

Solution:
The man is looking at a picture of his son.

• “The father of the person in the picture is my father’s son” means the person in the picture is the man’s son, because “my father’s son” refers to the man himself.

Example 54: The Light Bulb Puzzle (Revisited)

Question:
You are in a room with three light switches. Each switch controls one of three light bulbs in another room. You can only enter the room once to check the bulbs. How do you determine which switch controls which bulb?

Solution:

1. Turn on the first switch and leave it on for a few minutes to let the bulb heat up.
2. Turn off the first switch and turn on the second switch.
3. Immediately enter the room:
   • The bulb that is on is controlled by the second switch.
   • The bulb that is off but warm is controlled by the first switch.
   • The bulb that is off and cold is controlled by the third switch.

Example 55: The Mysterious Number Puzzle

Question:
What is the next number in the sequence: 2, 6, 12, 20, 30, ___?

Solution:
This sequence follows a pattern of adding consecutive even numbers to the previous term:

• 2 + 4 = 6
• 6 + 6 = 12
• 12 + 8 = 20
• 20 + 10 = 30
  So, the next number is 30 + 12 = 42.

Example 56: The Well and Rope Puzzle

Question:
A rope is tied to the top of a well. The rope is 30 feet long, and the well is 20 feet deep. Every time you pull the rope, you bring it up 5 feet. How many pulls does it take to get the rope completely out of the well?

Solution:
On the first pull, the rope comes up 5 feet, so now only 15 feet of rope is still in the well. On the second pull, 5 feet of rope is brought up, leaving 10 feet in the well. After the third pull, 5 feet is pulled up, leaving 5 feet in the well. On the fourth pull, you will pull the last 5 feet of rope, getting the rope completely out.

Thus, it will take 4 pulls to get the rope out of the well.

Example 57: The Missing Number Puzzle

Question:
What number comes next in this sequence: 3, 6, 9, 12, 15, ___?

Solution:
This is an arithmetic sequence where each number increases by 3.

• 3 + 3 = 6
• 6 + 3 = 9
• 9 + 3 = 12
• 12 + 3 = 15
  Thus, the next number in the sequence is 18.

Example 58: The Two Fathers Puzzle (Revisited)

Question:
A man and his son are sitting together. Both the man and the son are fathers and sons. How is this possible?

Solution:
The man is a father and also a son. He is a father to his son, and his own father is still alive. Therefore, the man is both a father and a son, and his son is also a son and a father (in the future). The key is the grandfather role.

Example 59: The Clock Puzzle

Question:
At exactly 3:15, what is the angle between the hour hand and the minute hand of a clock?

Solution:
The hour hand moves 360 degrees in 12 hours, so it moves 30 degrees per hour. At 3:00, the hour hand is at the 3 o’clock position (90 degrees).
At 3:15, the hour hand has moved a quarter of the way to the 4 o’clock position, which is 30 degrees/4 = 7.5 degrees.
The minute hand moves 360 degrees in 60 minutes, so at 15 minutes past the hour, it is at the 3 o’clock position (90 degrees).
Thus, the angle between the hour and minute hands is:
90 degrees (minute hand) – 97.5 degrees (hour hand) = 7.5 degrees.

Example 60: The Coin Flip Puzzle

Question:
You have 10 coins, each showing heads. You are blindfolded, and you must divide the coins into two groups with an equal number of heads. How do you do it?

Solution:

1. Divide the coins into two groups of 5 coins each.
2. Flip all the coins in one of the groups.
   Since you started with 10 heads in total, flipping the coins in one group will result in exactly 5 heads in each group.

Example 61: The Incomplete Calendar Puzzle

Question:
If January 1st of a certain year is a Monday, what day of the week will be the first day of the next year?

Solution:
Since the year has 365 days (in a non-leap year), we divide 365 by 7 (since there are 7 days in a week).
365 ÷ 7 = 52 weeks and 1 extra day.
This means that the first day of the next year will be 1 day ahead of Monday.
So, the first day of the next year will be Tuesday.

Example 62: The Coin Puzzle

Question:
You have 10 coins placed in a row. You are blindfolded, and you know that 5 of them are heads and 5 are tails. How can you separate the coins into two groups, each with the same number of heads?

Solution:

1. Separate the 10 coins into two groups of 5 coins each.
2. Flip all the coins in one of the groups.
   This works because, before flipping, if there are “X” heads in one group, there must be “5-X” heads in the other group.
   Flipping all the coins in the first group will change “X” heads to tails and “5-X” tails to heads, thus making both groups have the same number of heads.

Example 63: The Family Puzzle (Version 2)

Question:
A man has two daughters. One is 12 years old, and the other is half her age. How old will the older daughter be when the younger daughter turns 18?

Solution:
The older daughter is 12, and the younger daughter is half her age, so the younger daughter is 6 years old.
The age gap between the two daughters is 6 years.
So, when the younger daughter turns 18, the older daughter will be 18 + 6 = 24 years old.

Example 64: The Two Roads Puzzle

Question:
You are standing at a fork in the road. One road leads to a dangerous city, and the other leads to safety. There are two guards: one always tells the truth, and the other always lies. You do not know which guard is which. You can ask one question to one of the guards to determine which road leads to safety. What question do you ask?

Solution:
You should ask either guard: “If I were to ask the other guard which road leads to safety, which road would they point to?”

• If you ask the truthful guard, they will tell you the road that the liar would point to (the dangerous road).
• If you ask the lying guard, they will lie about what the truthful guard would say and point to the dangerous road. In both cases, the road they point to is the dangerous road. So, take the opposite road.

Example 65: The Vowel Puzzle

Question:
You have a five-letter word. When you remove the first and last letters, it becomes a four-letter word. When you remove the first letter of the new word, it becomes a three-letter word. When you remove the first letter of that word, it becomes a two-letter word. Finally, when you remove the first letter of that two-letter word, you are left with a one-letter word. What word is this?

Solution:
The word is “stone”.

• “Stone” → Remove the first and last letters → “ton”
• “Ton” → Remove the first letter → “on”
• “On” → Remove the first letter → “n”
  So, the word follows the pattern: stone → ton → on → n.

Also Read: Reasoning question on Number, Ranking and time sequence

Example 66: The Counterfeit Coin Puzzle

Question:
You have 12 coins, and one of them is counterfeit. The counterfeit coin is either heavier or lighter than the others, but you don’t know which. Using a balance scale, how can you find the counterfeit coin in just 3 weighings?

Solution:

1. Divide the 12 coins into three groups of 4 coins each.
2. Weigh two groups of 4 coins against each other.
   • If they balance, the counterfeit coin is in the third group of 4 coins.
   • If they don’t balance, the counterfeit coin is in the lighter or heavier group.
3. Take the group with the counterfeit coin and divide it into three groups of 1 coin each.
4. Weigh two of the coins against each other.
   • If they balance, the third coin is the counterfeit one.
   • If they don’t balance, the heavier or lighter coin is the counterfeit one.

Thus, you find the counterfeit coin in just 3 weighings.

Example 67: The Ladder Puzzle

Question:
You are climbing a ladder with 10 steps. Every time you climb 2 steps, you go 1 step back. How many steps will you need to climb to reach the top?

Solution:
For every 2 steps you climb, you go back 1 step. So, for every 3 steps (2 steps up and 1 step back), you effectively move 1 step up.
To reach the 10th step, you’ll need to perform this cycle 9 times (covering 9 steps). On the 10th step, you simply climb 2 steps without going back.
Thus, you will need to climb 19 steps in total.

Example 68: The Calendar Puzzle

Question:
What is the probability that your birthday will fall on a Monday, given that you were born on a random day of the year?

Solution:
Since there are 7 days in a week, and each day has an equal chance of being your birthday, the probability of your birthday falling on a Monday is simply 1/7 or approximately 14.29%.

Example 69: The Family Tree Puzzle

Question:
A man is looking at a picture. His friend asks, “Who are you looking at?” The man replies, “Brothers and sisters, I have none. But the father of the person in the picture is my father’s son.” Who is the person in the picture?

Solution:
The man is looking at a picture of his son.

• “The father of the person in the picture is my father’s son” means the person in the picture is the man’s son, because “my father’s son” refers to the man himself.

Example 70: The Train Puzzle

Question:
A train travels from Station A to Station B in 5 hours. If the train were to travel at twice its speed, how long would the journey take?

Solution:
If the speed doubles, the time taken will be halved.
So, if the train originally takes 5 hours, with double the speed, the journey would take 2.5 hours.

Example 71: The Reverse Number Puzzle

Question:
What number is equal to the sum of its digits when the digits are reversed?

Solution:
Let’s use the two-digit number AB, where A is the tens digit and B is the ones digit.
The number is represented as 10A+B10A + B10A+B.
When the digits are reversed, the number becomes 10B+A10B + A10B+A.
We need to find the number such that:
10A+B=10B+A10A + B = 10B + A10A+B=10B+A.
Solving for this, we get:
9A=9B9A = 9B9A=9B, which simplifies to:
A=BA = BA=B.
Thus, the number must be of the form AA (both digits are the same), and the number is 11.

Example 72: The Egg Puzzle

Question:
You have two eggs and a 100-story building. You want to find the highest floor from which an egg can be dropped without breaking, but you can only drop the eggs from different floors a limited number of times. What is the minimum number of drops needed?

Solution:
To minimize the number of drops, you need to start dropping the eggs from higher floors and gradually lower the starting floor after each drop. The optimal strategy is to start at floor 14 and move down by one less floor after each drop (e.g., 14, then 27, then 39, etc.).
This way, you minimize the number of drops in the worst-case scenario, and the minimum number of drops is 14.

Example 73: The Three Switch Puzzle

Question:
You have three switches in one room. Each switch controls one of three light bulbs in another room. You can only enter the room once to check the bulbs. How do you determine which switch controls which bulb?

Solution:

1. Turn on the first switch and leave it on for a few minutes to let the bulb heat up.
2. Turn off the first switch and turn on the second switch.
3. Immediately enter the room:
   • The bulb that is on is controlled by the second switch.
   • The bulb that is off but warm is controlled by the first switch.
   • The bulb that is off and cold is controlled by the third switch.

Example 74: The 100 Prisoners Puzzle

Question:
100 prisoners are lined up in a row, each with a hat on their head. Each prisoner can see the hats of the others but not their own. The colors of the hats are either black or white. Starting from the back of the line, each prisoner must guess the color of their own hat. If they guess correctly, they are set free. What is the best strategy to ensure that the maximum number of prisoners guess their hats correctly?

Solution:
The strategy involves the first prisoner (who can see all the hats) announcing a color based on the parity of the number of black hats he sees. If the number of black hats is even, the first prisoner says “white,” and if it is odd, he says “black.” This gives the remaining prisoners a way to deduce their own hats based on the number of black hats they can see.
With this strategy, at least 99 prisoners will be guaranteed to guess correctly, while the first prisoner may have to guess.

Example 75: The Light Bulb and Switch Puzzle (Revisited)

Question:
You have 8 light bulbs, all turned off. In the next room, there are 8 switches, each connected to one light bulb. You can only go into the room with the bulbs once. How can you determine which switch controls which bulb?

Solution:

1. Turn on 3 switches and leave them on for a while.
2. After a few minutes, turn off one of the switches and turn on the 4th switch.
3. Go into the room:
   • The bulb that is on is controlled by the 4th switch.
   • The bulb that is off but warm is controlled by the first switch (the one that was on for a while).
   • The bulb that is off and cold is controlled by the second switch (the one that was turned off after being on for a while).
   • The bulb that is off and cold is controlled by the third switch (the one that was never turned on).

Example 76: The Riddle of the River Crossing

Question:
A man has a wolf, a goat, and a cabbage. He needs to cross a river, but his boat can only carry himself and one item at a time. If he leaves the wolf and the goat together, the wolf will eat the goat. If he leaves the goat and the cabbage together, the goat will eat the cabbage. How does he get all three across the river safely?

Solution:

1. Take the goat across first.
2. Go back to the original side alone.
3. Take the wolf across.
4. Bring the goat back with you.
5. Take the cabbage across.
6. Go back to the original side alone.
7. Finally, take the goat across.

This way, all three (the wolf, the goat, and the cabbage) make it safely across the river.

Example 77: The Two Sons Puzzle

Question:
A man has two sons. The older son is 4 years older than the younger son. The sum of their ages is 24. How old are the sons?

Solution:
Let the younger son’s age be xxx.
The older son’s age is x+4x + 4x+4.
The sum of their ages is 24, so:
x+(x+4)=24x + (x + 4) = 24x+(x+4)=24
2x+4=242x + 4 = 242x+4=24
2x=202x = 202x=20
x=10x = 10x=10
So, the younger son is 10 years old, and the older son is 10+4=1410 + 4 = 1410+4=14 years old.

Example 78: The 5 Liter Jug Puzzle

Question:
You have two jugs: one can hold 5 liters, and the other can hold 3 liters. You need to measure exactly 4 liters of water. How do you do it?

Solution:

1. Fill the 5-liter jug completely.
2. Pour water from the 5-liter jug into the 3-liter jug until the 3-liter jug is full. This leaves 2 liters in the 5-liter jug.
3. Empty the 3-liter jug.
4. Pour the remaining 2 liters from the 5-liter jug into the 3-liter jug.
5. Fill the 5-liter jug again.
6. Pour water from the 5-liter jug into the 3-liter jug until the 3-liter jug is full (it already has 2 liters).
7. This leaves exactly 4 liters in the 5-liter jug.

Example 79: The Candies Puzzle

Question:
You have 8 identical-looking candies, and one of them is poisoned. You have a balance scale to use, and you can only use it two times. How do you determine which candy is poisoned?

Solution:

1. Divide the 8 candies into three groups: 3 candies in the first group, 3 candies in the second group, and 2 candies in the third group.
2. Weigh the first two groups (3 candies vs. 3 candies):
   • If they balance, the poisoned candy must be in the group of 2 candies.
   • If they do not balance, the poisoned candy must be in the heavier group.
3. Now, with the group of 3 or 2 candies, weigh two of them against each other.
   • If they balance, the third candy is poisoned.
   • If they do not balance, the heavier candy is poisoned.

Thus, you find the poisoned candy in 2 weighings.

Example 80: The Wolf, Goat, and Cabbage Puzzle (Variation)

Question:
A man has a wolf, a goat, and a cabbage, and he needs to cross a river with these items. He can only take one item at a time. If he leaves the wolf and the goat together, the wolf will eat the goat. If he leaves the goat and the cabbage together, the goat will eat the cabbage. How does he get all three across the river without any of them being eaten?

Solution:

1. First, take the goat across the river and leave it on the other side.
2. Go back alone and take the wolf across the river.
3. Bring the goat back with you.
4. Take the cabbage across the river.
5. Go back alone to get the goat.
6. Finally, take the goat across the river again.

This way, no one gets eaten.

Example 81: The Age Puzzle

Question:
Five years ago, Jane was 3 times as old as her daughter. In 10 years, Jane will be 2 times as old as her daughter. How old are Jane and her daughter now?

Solution:
Let Jane’s current age be jjj, and her daughter’s current age be ddd.
From the first condition (5 years ago):
j−5=3(d−5)j – 5 = 3(d – 5)j−5=3(d−5)
This simplifies to:
j−5=3d−15j – 5 = 3d – 15j−5=3d−15
j=3d−10j = 3d – 10j=3d−10 (Equation 1)

From the second condition (in 10 years):
j+10=2(d+10)j + 10 = 2(d + 10)j+10=2(d+10)
This simplifies to:
j+10=2d+20j + 10 = 2d + 20j+10=2d+20
j=2d+10j = 2d + 10j=2d+10 (Equation 2)

Now, equate the two equations:
3d−10=2d+103d – 10 = 2d + 103d−10=2d+10
d=20d = 20d=20
Substitute d=20d = 20d=20 into Equation 2:
j=2(20)+10=50j = 2(20) + 10 = 50j=2(20)+10=50

Thus, Jane is 50 years old, and her daughter is 20 years old.

Example 82: The Missing Dollar Puzzle

Question:
Three friends go to a hotel and check into a room. The cost of the room is $30. Each friend contributes $10, giving a total of $30. Later, the hotel manager realizes there’s a mistake, and the room should have only cost $25. The manager gives $5 to the bellboy and asks him to return it to the three friends. The bellboy decides to keep $2 for himself and gives $1 back to each of the three friends. Now, each friend has paid $9 (3 × $9 = $27) and the bellboy kept $2, which adds up to $29. Where is the missing dollar?

Solution:
The trick in this puzzle is in the way the question is phrased. The $27 (paid by the friends) includes the $25 for the room and the $2 that the bellboy kept. The remaining dollar is accounted for in the $25 cost of the room. There is no missing dollar. The confusion arises from adding the bellboy’s $2 to the $27 instead of subtracting it.

Example 83: The Paradoxical Journey Puzzle

Question:
You are on a road trip and driving at a speed of 60 miles per hour. You are 120 miles away from your destination. How long will it take you to reach your destination?

Solution:
At 60 miles per hour, you will cover 60 miles in one hour. Since you are 120 miles away, the time required to reach your destination is:
12060=2 hours\frac{120}{60} = 2 \text{ hours}60120​=2 hours
So, it will take you 2 hours to reach your destination.

Example 84: The Odd One Out Puzzle

Question:
Which number is the odd one out: 3, 6, 9, 12, 15, 18, 20, 21?

Solution:
The odd one out is 20, because it is the only number in the sequence that is not a multiple of 3.

Example 85: The Three Friends Puzzle

Question:
Three friends, Alice, Bob, and Carol, are walking in a park. Alice is facing north, Bob is facing south, and Carol is facing east. They all turn around, but none of them turns more than 180 degrees. After they turn, where are they facing?

Solution:

1. Alice turns 180 degrees from north, so she is now facing south.
2. Bob turns 180 degrees from south, so he is now facing north.
3. Carol turns 180 degrees from east, so she is now facing west.

Thus, after turning, Alice is facing south, Bob is facing north, and Carol is facing west.

Example 86: The Brothers Puzzle

Question:
A man has two brothers. He is 4 years older than his younger brother, and the sum of their ages is 30. How old are the brothers?

Solution:
Let the younger brother’s age be xxx.
The man is 4 years older, so his age is x+4x + 4x+4.
The sum of their ages is 30:x+(x+4)=30x + (x + 4) = 30x+(x+4)=30 2x+4=302x + 4 = 302x+4=30 2x=262x = 262x=26 x=13x = 13x=13

So, the younger brother is 13, and the man is 13+4=1713 + 4 = 1713+4=17.

Example 87: The Coin Puzzle (Variation)

Question:
You have two coins. One is a normal coin, and the other is a fake coin. The fake coin is either heavier or lighter than the normal coin. How can you identify the fake coin using just one weighing on a balance scale?

Solution:

1. Place the two coins on the balance scale, one on each side.
2. If the scale tips to one side, the heavier or lighter coin is the fake one.
3. If the scale is balanced, the coin you didn’t place on the scale is the fake one.

Thus, you can identify the fake coin in one weighing.

Example 88: The River Crossing with the Fox, Goose, and Bag of Beans

Question:
A farmer is crossing a river with a fox, a goose, and a bag of beans. He has a boat, but it can only carry him and one item at a time. If he leaves the fox with the goose, the fox will eat the goose. If he leaves the goose with the beans, the goose will eat the beans. How can the farmer get all three across the river safely?

Solution:

1. Take the goose across the river first and leave it on the other side.
2. Go back and take the fox across.
3. Bring the goose back with you.
4. Take the beans across the river.
5. Go back and take the goose across.

By following this sequence, the farmer will safely get the fox, goose, and beans across the river.

Example 89: The Age Difference Puzzle

Question:
The sum of the ages of a mother and her daughter is 50. The mother is 4 times as old as her daughter. How old are the mother and the daughter?

Solution:
Let the daughter’s age be xxx.
The mother’s age is 4x4x4x.
The sum of their ages is 50:x+4x=50x + 4x = 50x+4x=50 5x=505x = 505x=50 x=10x = 10x=10

So, the daughter is 10 years old, and the mother is 4×10=404 \times 10 = 404×10=40 years old.

Example 90: The Coin Puzzle with Weighing (3 Coins)

Question:
You have three identical-looking coins, but one of them is slightly heavier than the other two. How can you find the heavier coin in just one weighing on a balance scale?

Solution:

1. Place two of the coins on the balance scale, one on each side.
2. If the scale tips to one side, the heavier coin is on the heavier side.
3. If the scale is balanced, the third coin (the one not on the scale) is the heavier one.

Thus, you can find the heavier coin in one weighing.

Example 91: The Clock Puzzle

Question:
At 12:00 PM, the hands of the clock are at the same position. How many times do the hands of the clock overlap between 12:00 PM and 1:00 PM?

Solution:
The hands of the clock overlap exactly once between 12:00 PM and 1:00 PM. Although the minute hand completes a full revolution every hour, it overtakes the hour hand once in that period. This occurs approximately at 12:32.

Example 92: The Weighing Puzzle (Group of 8)

Question:
You have 8 coins, and one of them is either heavier or lighter than the others. You are allowed to use a balance scale only twice. How can you find the odd coin and determine whether it is heavier or lighter?

Solution:

1. Divide the 8 coins into three groups: two groups of 3 coins each and one group of 2 coins.
2. Weigh the two groups of 3 coins against each other.
   • If they balance, the odd coin is in the group of 2 coins.
   • If they don’t balance, the odd coin is in the group that is heavier or lighter.
3. Take the group with the odd coin and weigh two coins against each other.
   • If they balance, the remaining coin is the odd one.
   • If they don’t balance, the heavier or lighter coin is the odd one.

Thus, you can find the odd coin and determine whether it is heavier or lighter in just two weighings.

Example 93: The Mystery Box Puzzle

Question:
You have 10 boxes. Each box contains 10 items, but one box contains a weight that is 1 gram heavier than the rest. You have a scale that can only be used once. How can you find the box with the heavier items?

Solution:

1. Number the boxes from 1 to 10.
2. From each box, take a number of items equal to the number of the box. For example, from box 1, take 1 item; from box 2, take 2 items; and so on.
3. Weigh all the selected items together on the scale.
4. The weight of the items should be 10×10=10010 \times 10 = 10010×10=100 grams if all the items weigh the same. However, if one box has heavier items, the weight will be more than 100 grams by a certain amount.
   • If the total weight is 101 grams, the heavier box is Box 1.
   • If the total weight is 102 grams, the heavier box is Box 2, and so on.

This way, you can find the heavier box in one weighing.

Example 94: The Sealed Envelope Puzzle

Question:
You have 5 sealed envelopes, each containing a different amount of money. You know that:

• The first envelope has more money than the second.
• The second envelope has more money than the third.
• The fourth envelope has less money than the fifth.

Can you determine the exact order of the envelopes in terms of the amount of money without opening them?

Solution:
From the clues:

• Envelope 1 has more than Envelope 2.
• Envelope 2 has more than Envelope 3.
• Envelope 4 has less than Envelope 5.

Thus, the order must be:

1. Envelope 1
2. Envelope 2
3. Envelope 3
4. Envelope 5
5. Envelope 4

So, the correct order is: 1, 2, 3, 5, 4.

Example 95: The Bag of Marbles Puzzle

Question:
You have a bag containing 100 marbles. 99 of them are red, and 1 is blue. You are blindfolded, and you are allowed to pick out marbles one by one without looking. What is the minimum number of marbles you must take out to be certain that you have at least 2 marbles of the same color?

Solution:
The worst-case scenario is that you pick one red marble and the blue marble. The next marble you pick will certainly be red because there are 99 red marbles. Thus, the minimum number of marbles you need to pick is 3.

Example 96: The Brothers and Sisters Puzzle

Question:
A father has 6 children. Each of his children has a different number of brothers and sisters. How many brothers and how many sisters does each child have?

Solution:
Let’s assume the children are labeled A, B, C, D, E, and F. Each child has 5 siblings, but the number of brothers and sisters each child has will vary.

• A has 5 siblings: 4 brothers and 1 sister.
• B has 5 siblings: 4 brothers and 1 sister.
• C has 5 siblings: 4 brothers and 1 sister.
• D has 5 siblings: 4 brothers and 1 sister.
• E has 5 siblings: 4 brothers and 1 sister.
• F has 5 siblings: 4 brothers and 1 sister.

Thus, each child has 4 brothers and 1 sister.

Example 97: The Four People Puzzle

Question:
Four people are standing in a line. The first person is taller than the second person, but shorter than the third person. The third person is shorter than the fourth person. Who is the shortest person?

Solution:
From the clues:

1. The first person is taller than the second person but shorter than the third person. This means the order is: Second person, First person, Third person.
2. The third person is shorter than the fourth person. So the order is: Second person, First person, Third person, Fourth person.

Therefore, the second person is the shortest.

Example 98: The Candle Puzzle

Question:
You have a candle that burns for exactly 60 minutes. However, the candle is not uniform and burns unevenly, meaning it may burn faster at one end and slower at the other. How can you use this candle to measure exactly 45 minutes?

Solution:

1. Light the candle at both ends simultaneously.
2. Since the candle burns unevenly, lighting both ends will cause it to burn completely in 30 minutes (because burning from both ends effectively halves the time it takes).
3. Once the candle is halfway burned (after 30 minutes), immediately light the second end of the candle.
4. The candle will burn completely in 15 minutes after lighting the second end.
5. So, you have measured 45 minutes by combining the 30 minutes and 15 minutes.

Example 99: The Box of Balls Puzzle

Question:
There are 3 boxes. One contains only red balls, one contains only green balls, and one contains both red and green balls. The boxes are labeled incorrectly. You are allowed to pick one ball from one box. How can you label the boxes correctly?

Solution:

1. Pick one ball from the box labeled “both red and green.” Since the label is incorrect, the box must contain only one color, either red or green.
   • If you pick a red ball, this box contains only red balls.
   • If you pick a green ball, this box contains only green balls.
2. Now, you know the content of the box you picked from. For the remaining two boxes, the labels are also incorrect, so the box labeled “red” must contain both colors, and the box labeled “green” must contain the other single color.

Thus, you can label the boxes correctly by drawing just one ball.

Example 100: The Two Doors Puzzle

Question:
You are in a room with two doors. One leads to freedom, and the other leads to certain death. There are two guards, one in front of each door. One guard always tells the truth, and the other always lies. You do not know which guard is which. You may ask only one question. What question do you ask to determine which door leads to freedom?

Solution:
Ask either guard, “If I were to ask the other guard which door leads to freedom, which door would he point to?”

• If you ask the truthful guard, they will tell you the door that leads to death (because the lying guard would point to the wrong door).
• If you ask the lying guard, they will lie about what the truthful guard would say and point to the wrong door.

In both cases, you will be directed to the door that leads to death. So, you should choose the other door.

Example 101: The Egg Puzzle

Question:
You have 3 eggs and a 100-story building. You need to determine the highest floor from which you can drop an egg without it breaking. You can drop the eggs as many times as you need, but once an egg breaks, it cannot be used again. What is the minimum number of drops needed to find the highest safe floor?

Solution:
The best strategy is to drop the eggs in a way that balances the worst-case scenario. Start by dropping the first egg from the 14th floor, then increase the floor by one less floor each time (i.e., the next drop is from the 27th floor, then 39th, and so on). If the first egg breaks, you can use the second egg to drop from the previous floor and work your way up from there.

This minimizes the number of drops in the worst case, and you can determine the highest safe floor in 14 drops.

Example 102: The Light Bulb Puzzle

Question:
You are in a room with three light bulbs, each controlled by a separate switch outside the room. You can flip the switches on and off as many times as you want, but you can only enter the room once to check the bulbs. How do you figure out which switch controls which bulb?

Solution:

1. Turn on the first switch and leave it on for a few minutes.
2. After a few minutes, turn off the first switch and immediately turn on the second switch.
3. Enter the room:
   • The bulb that is on corresponds to the second switch.
   • The bulb that is off but warm corresponds to the first switch.
   • The bulb that is off and cold corresponds to the third switch.

Thus, you can determine which switch controls which bulb in one trip.

Example 103: The 3 People Puzzle

Question:
Three people (A, B, and C) are standing in a line. A can see B and C, B can only see C, and C can’t see anyone. Each person is wearing a hat that is either black or white, and they know that at least one person has a black hat. Each person must guess the color of their own hat. What is the minimum number of guesses needed for all three to guess their hat colors correctly?

Solution:

1. A, who can see B and C’s hats, knows that at least one person is wearing a black hat. If A sees two white hats, they will immediately know they are wearing a black hat.
2. If A sees one black hat and one white hat, A cannot immediately determine their own hat color.
3. B, seeing only C’s hat, will guess based on what A knows. If B sees a white hat on C, B can conclude that A must be wearing a black hat, and if B sees a black hat on C, B will guess that their own hat is white.

By this reasoning, all three people can guess their hat colors in one or two guesses.

Example 104: The Cipher Puzzle

Question:
A secret message reads:
“GUR YVQR QNQ, ABJ”

What is the decoded message?

Solution:
This is an example of the ROT13 cipher, where each letter is shifted by 13 places in the alphabet. Decoding it, we get:
“THE LID DEC, NOW”

The decoded message is “THE OLD CODE, NOW.”

Example 105: The Two Sisters Puzzle

Question:
Two sisters, Alice and Beth, are 10 years apart in age. In 6 years, Alice will be twice as old as Beth. How old are Alice and Beth now?

Solution:
Let Alice’s current age be xxx, and Beth’s current age is x−10x – 10x−10.
In 6 years, Alice’s age will be x+6x + 6x+6, and Beth’s age will be (x−10)+6=x−4(x – 10) + 6 = x – 4(x−10)+6=x−4.
We are told that in 6 years, Alice will be twice as old as Beth:x+6=2(x−4)x + 6 = 2(x – 4)x+6=2(x−4)

Solving for xxx:x+6=2x−8x + 6 = 2x – 8x+6=2x−8 6+8=2x−x6 + 8 = 2x – x6+8=2x−x x=14x = 14x=14

So, Alice is 14 years old, and Beth is 4 years old.

Example 106: The Coin Flip Puzzle

Question:
You flip a coin 100 times. What is the minimum number of heads you are guaranteed to get?

Solution:
In the worst case, all the flips result in tails. So, if the coin flips are all tails, the minimum number of heads is 0.

However, in terms of probability or expectation, you can expect to get about 50 heads on average.

Example 107: The Hat and the Shoes Puzzle

Question:
A group of 100 people are each wearing a hat. The hats are either black or white, and each person can see everyone else’s hat but not their own. Each person must simultaneously guess the color of their own hat. If the guess is correct, they live; if it’s wrong, they die. What strategy can guarantee that at least 99 of the 100 people will survive?

Solution:
The strategy involves a binary method:

1. The first person (Person 1) guesses based on the number of black hats they see. They will guess “black” if they see an odd number of black hats and “white” if they see an even number of black hats.
2. Every subsequent person will base their guess on the previous guesses and the number of black hats they can see. By using the parity of the black hats seen, everyone except the first person can guarantee that their guess is correct.

In this way, at least 99 people will survive, and only the first person might guess incorrectly.

Example 108: The Birthday Paradox Puzzle

Question:
In a room of 23 people, what is the probability that at least two people share the same birthday?

Solution:
The probability that no two people share the same birthday is calculated by multiplying the probabilities that each successive person does not share a birthday with anyone before them:P(no shared birthday)=365365×364365×363365×⋯×343365P(\text{no shared birthday}) = \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \cdots \times \frac{343}{365}P(no shared birthday)=365365​×365364​×365363​×⋯×365343​

This gives approximately a 49.3% chance that no one shares a birthday. Thus, the probability that at least two people share a birthday is about:1 – 0.493 = 0.507 \quad \text{or approximately 50.7%}.

Example 109: The Family Puzzle

Question:
A man is looking at a picture of someone. His friend asks, “Who are you looking at?” The man replies, “Brothers and sisters, I have none. But the father of the person in the picture is my father’s son.” Who is the man looking at?

Solution:
The key clue is: “The father of the person in the picture is my father’s son.” The man is referring to himself as his father’s son. Thus, the person in the picture is his son.

Example 110: The Elevator Puzzle

Question:
A man lives on the 10th floor of a building. Every morning, he takes the elevator to the ground floor and leaves for work. In the evening, he takes the elevator back up. However, when returning, he only takes the elevator to the 7th floor and walks the rest of the way up. Why?

Solution:
The man is short and cannot reach the button for the 10th floor. In the morning, he can press the ground floor button, but when returning, he can only reach the button for the 7th floor, so he walks the remaining 3 floors.

Example 111: The Candles Puzzle

Question:
You have 2 candles. Each candle burns for exactly 60 minutes, but they burn unevenly (not at a uniform rate). How can you use these 2 candles to measure 45 minutes?

Solution:

1. Light both candles at the same time, but light one from both ends and the other from just one end.
2. The candle that is lit from both ends will burn completely in 30 minutes.
3. Once the first candle is completely burned (after 30 minutes), light the second end of the other candle.
4. The second candle will burn completely in the next 15 minutes.
   Thus, the total time measured will be 45 minutes.

Example 112: The Pirates and the Treasure Puzzle

Question:
Five pirates discover a treasure of 100 gold coins. They decide to divide the treasure according to the following rules:

1. The oldest pirate proposes a division of the treasure.
2. The pirates then vote on the proposal.
3. If more than half of the pirates agree, the treasure is divided as proposed. If not, the oldest pirate is thrown overboard, and the next oldest pirate proposes a new division.
4. Pirates are very logical and want to stay alive, but they also want to maximize the amount of gold they get.

How should the oldest pirate propose to divide the coins to ensure his survival?

Solution:
The key to this puzzle is knowing how the pirates will vote. The pirates will always prefer to throw someone overboard if they can get more gold, but they are also very logical and want to avoid dying.

• If there is only 1 pirate left, he keeps all 100 coins.
• If there are 2 pirates left, the oldest one keeps all 100 coins (since the younger pirate has no vote power).
• If there are 3 pirates left, the oldest pirate needs the support of at least 1 pirate. The oldest pirate gives 1 coin to the youngest pirate (who has no voting power otherwise) and keeps 99 coins.
• If there are 4 pirates left, the oldest pirate knows that the youngest pirate can be persuaded to vote for the proposal since he would get nothing if the oldest pirate is thrown overboard. The oldest pirate keeps 98 coins, gives 1 coin to the youngest pirate, and 1 coin to the second-youngest pirate.
• If there are 5 pirates left, the oldest pirate gives 1 coin to the third-youngest pirate (who would otherwise get nothing), 1 coin to the youngest pirate, and keeps 98 coins.

Thus, the oldest pirate should propose to keep 98 coins, give 1 coin to the third-youngest pirate, and 1 coin to the youngest pirate. The second-youngest pirate would not get anything in this case, but the vote will be in favor because the other pirates prefer to get something rather than nothing.

Example 113: The 100 Prisoners Puzzle

Question:
100 prisoners are each given a number from 1 to 100. They are each placed in a room with a drawer in front of them, containing a random arrangement of their numbers. The prisoners are allowed to open up to 50 drawers, and they must try to find their own number. If all prisoners find their own number, they are set free. If even one prisoner fails to find their number, they all die. What is the best strategy for the prisoners?

Solution:
The optimal strategy is for each prisoner to open the drawer corresponding to their number, then open the drawer corresponding to the number found in the first drawer, and so on, following a chain of numbers. Each prisoner only follows the chain for 50 steps, and the chance of all prisoners surviving is about 31%. This strategy works because the chain will either loop back on itself within 50 steps or continue long enough to allow all prisoners to find their numbers.

Example 114: The Meeting Room Puzzle

Question:
Five friends (A, B, C, D, and E) are meeting in a room. They all have different preferences about the seat they want to sit in:

• A doesn’t want to sit next to B.
• B doesn’t want to sit next to C.
• C doesn’t want to sit next to D.
• D doesn’t want to sit next to E.
• E doesn’t want to sit next to A.

How can they sit so that no one is sitting next to the person they don’t want to sit next to?

Solution:
A possible seating arrangement is:

• A, D, B, E, C

This arrangement satisfies all the conditions:

• A is not next to B.
• B is not next to C.
• C is not next to D.
• D is not next to E.
• E is not next to A.

Example 115: The Colored Balls Puzzle

Question:
You have 3 bags:

• Bag 1 contains 2 red balls.
• Bag 2 contains 2 green balls.
• Bag 3 contains 1 red ball and 1 green ball.

Each bag is mislabeled. You are allowed to pick one ball from one bag. How can you figure out the correct labels of the bags?

Solution:
Pick a ball from the bag labeled “Mixed” (which must contain either 2 red balls or 2 green balls because it is mislabeled).

• If you pick a red ball, then the “Mixed” bag contains 2 red balls.
• If you pick a green ball, then the “Mixed” bag contains 2 green balls.

After finding the contents of the “Mixed” bag, you can deduce the contents of the other two bags:

• The bag labeled “Red” must contain 1 red ball and 1 green ball (since the “Mixed” bag cannot contain both red and green balls).
• The bag labeled “Green” must contain 2 green balls if the “Mixed” bag contained 2 red balls, or vice versa.

Example 116: The Stairs Puzzle

Question:
A person can climb a staircase in two ways: by taking 1 step at a time or by taking 2 steps at a time. How many ways can the person climb a staircase with 5 steps?

Solution:
To find the total number of ways to climb the staircase, consider the following:

• If the person takes 1 step on the first move, they must climb 4 more steps.
• If the person takes 2 steps on the first move, they must climb 3 more steps.

Thus, the total number of ways to climb 5 steps is the sum of the ways to climb 4 steps and the ways to climb 3 steps.
For 4 steps, there are 5 ways, and for 3 steps, there are 3 ways.
So, the total number of ways is 8.

Example 117: The Age Puzzle

Question:
A mother is 30 years older than her son. In 6 years, the mother’s age will be 5 times her son’s age. How old are the mother and the son now?

Solution:
Let the son’s age be xxx and the mother’s age be x+30x + 30x+30.
In 6 years, the mother’s age will be x+30+6=x+36x + 30 + 6 = x + 36x+30+6=x+36, and the son’s age will be x+6x + 6x+6.
We are told that the mother’s age will be 5 times the son’s age in 6 years:x+36=5(x+6)x + 36 = 5(x + 6)x+36=5(x+6)

Solving for xxx:x+36=5x+30x + 36 = 5x + 30x+36=5x+30 36−30=5x−x36 – 30 = 5x – x36−30=5x−x 6=4×6 = 4×6=4x x=1.5x = 1.5x=1.5

Thus, the son is 1.5 years old, and the mother is 31.5 years old.

Example 118: The Coin Puzzle

Question:
You have a coin that is weighted unevenly, but you don’t know whether it is heavier or lighter. You are allowed to use a balance scale, and you can use it up to 3 times. How can you determine whether the coin is heavier or lighter?

Solution:

1. Divide the 3 coins into two groups of 2 coins.
2. Weigh one group against the other.
   • If they balance, the unweighed coin is the odd one out, and you can determine if it’s heavier or lighter by simply testing it against one of the regular coins.
   • If they don’t balance, you’ll know which group contains the odd coin, and you can weigh one coin against the other to determine which is heavier or lighter.

This method ensures you can find the odd coin in 3 weighings.

Example 119: The Rope Puzzle

Question:
You have two ropes. Each rope has the property that if you light it at one end, it will take exactly 60 minutes to burn completely. However, the ropes don’t burn evenly — they may burn quickly at some parts and slowly at others. How can you measure exactly 45 minutes using these two ropes?

Solution:

1. Light the first rope at both ends and the second rope at one end at the same time.
2. The first rope will burn completely in 30 minutes because lighting it at both ends halves the burning time.
3. After 30 minutes, the first rope will be completely burnt. Now immediately light the second end of the second rope.
4. The second rope will take another 15 minutes to burn completely from both ends, so you will have a total of 45 minutes when the second rope finishes burning.

Example 120: The Age Difference Puzzle

Question:
A father is 30 years older than his son. In 10 years, the father will be three times as old as the son. How old are they now?

Solution:
Let the son’s age be xxx and the father’s age be x+30x + 30x+30.
In 10 years, the father’s age will be (x+30)+10=x+40(x + 30) + 10 = x + 40(x+30)+10=x+40, and the son’s age will be x+10x + 10x+10.
We are told that the father will be three times the son’s age in 10 years:x+40=3(x+10)x + 40 = 3(x + 10)x+40=3(x+10)

Solving for xxx:x+40=3x+30x + 40 = 3x + 30x+40=3x+30 40−30=3x−x40 – 30 = 3x – x40−30=3x−x 10=2×10 = 2×10=2x x=5x = 5x=5

So, the son is 5 years old, and the father is 35 years old.

Example 121: The River Crossing Puzzle

Question:
A farmer needs to cross a river with a wolf, a goat, and a cabbage. He has a boat that can only carry him and one other item. If left alone, the wolf will eat the goat, and the goat will eat the cabbage. How does the farmer get everything across safely?

Solution:

1. The farmer takes the goat across the river first and leaves it on the other side.
2. Then he goes back alone and takes the wolf across the river.
3. He leaves the wolf on the other side but takes the goat back with him.
4. He leaves the goat on the starting side and takes the cabbage across the river.
5. Finally, he returns to the starting side and takes the goat across.

By following this method, everything is successfully transported across without anything being eaten.

Example 122: The Prisoners and the Hats Puzzle

Question:
100 prisoners are lined up in a row, and each one is wearing a hat that is either black or white. The prisoners can see the hats of the people in front of them, but not their own hat or the hats of those behind them. Starting with the last prisoner (who can see all the hats in front of him), each prisoner must say “black” or “white” to indicate the color of their own hat. If they guess their hat color correctly, they survive; if not, they die. What is the strategy that guarantees at least 99 prisoners will survive?

Solution:
The last prisoner (the one who can see everyone else’s hats) will say “black” or “white” based on the parity of the number of black hats he sees in front of him.

1. If the number of black hats in front of him is even, he says “black.” If it’s odd, he says “white.”
2. Each subsequent prisoner will use the information from the previous guesses and the number of black hats they see in front of them to deduce the color of their own hat.

This strategy guarantees that at least 99 prisoners will survive, because only the last prisoner’s guess is uncertain, but the others can figure out their hat colors based on previous answers and the parity rule.

Example 123: The Three Switches Puzzle

Question:
You are in a room with 3 light switches, all in the off position. Each switch controls a separate light bulb in another room. You cannot see the bulbs from where you are. How can you determine which switch controls which bulb if you can only go into the other room once?

Solution:

1. Turn on the first switch and leave it on for a few minutes.
2. Turn off the first switch, and turn on the second switch.
3. Immediately go into the other room:
   • The light bulb that is on corresponds to the second switch.
   • The light bulb that is off but warm corresponds to the first switch.
   • The light bulb that is off and cold corresponds to the third switch.

Example 124: The Clock Puzzle

Question:
A clock shows the time as 3:15. How many times will the minute and hour hands be overlapping between 3:00 and 4:00?

Solution:
The hands overlap approximately once every hour, but because the hour hand is moving as well, the overlap between 3:00 and 4:00 will occur slightly after 3:15.
Thus, the hands will overlap once, and that will be at approximately 3:16:21.

Example 125: The Colored Marbles Puzzle

Question:
You have 3 jars. One jar is filled with only red marbles, another with only blue marbles, and the third with a mix of both red and blue marbles. Each jar is incorrectly labeled. How can you pick one marble from one jar to determine the correct labeling of all the jars?

Solution:

1. Pick a marble from the jar labeled “Mixed.”
   Since all jars are incorrectly labeled, the jar labeled “Mixed” must either be the red-only or blue-only jar.
2. If you pick a red marble, the jar must be the red-only jar, and the other two jars must be switched accordingly.
3. If you pick a blue marble, the jar must be the blue-only jar, and the remaining jars will be correctly identified.

Example 126: The Family Puzzle

Question:
A family has two parents and four children. The four children are the same age. How is this possible?

Solution:
The four children are triplets and one other child. The two parents are the mother and father, and the four children are the triplets and one more child who is the same age as the triplets.

Example 127: The Letter Puzzle

Question:
You are given a sequence of letters: C, O, D, E, C, O, D, E. What is the next letter in the sequence?

Solution:
The sequence repeats itself after every 4 letters: CODE. The next letter is C.

Example 128: The Two Doors Puzzle

Question:
You are in a room with two doors: one leads to freedom, and the other leads to certain death. You don’t know which is which. In front of each door, there is a guard. One guard always tells the truth, and the other always lies, but you don’t know which is which. You can ask one guard one question. What question do you ask to determine which door leads to freedom?

Solution:
Ask either guard: “If I were to ask the other guard which door leads to freedom, which door would they point to?”

• The truthful guard will point to the door of death because the liar would point to it.
• The liar will also point to the door of death because the truthful guard would point to it.

Thus, you should choose the opposite door.

All these type of Verbal Reasoning question on puzzles are helping for preparing competitive exam

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