**Puzzle 20:**A group of Englishmen are travelling on a German plane, piloted by an Norwegian, bound for Finland. If the plane was to crash in Sweden, where would the survivors be buried?

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You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.

What is the least number of races you can conduct to figure out which 3 horses are fastest?

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**Solution: **You need to conduct 7 races.

**Explanation: **

First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on.

We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races).

This table shows our remaining horses:

A1 | B1 | C1 | D1 | E1 |

A2 | B2 | C2 | D2 | E2 |

A3 | B3 | C3 | D3 | E3 |

For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last).

We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1).

Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3).

Let's look at our updated table, having removed these horses that can't be in the top 3:

A1 | B1 | C1 |

A2 | B2 | C2 |

A3 | B3 | C3 |

We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table:

A1 | B1 | C1 |

A2 | B2 | |

A3 |

We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.

One day, Bruce is in his cell talking to one of his cellmates, Steve.

"I really want to smoke 5 cigarettes today, but all I have are these 10 cigarette butts," Bruce tells Steve. "I'm not sure that will be enough."

"Why don't you borrow some of Tom's cigarette butts?" asks Steve, pointing over to a small pile of cigarette butts on the bed of their third cellmate, Tom, who is out for the day on a community service project.

"I can't," Bruce says. "Tom always counts exactly how many cigarette butts are in his pile, and he'd probably kill me if he noticed that I had taken any."

However, after thinking for a while, Bruce figures out a way that he can smoke 5 cigarettes without angering Tom. What is his plan?

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**Solution:** Bruce takes 9 of his 10 cigarette butts and turns them into 3 cigarettes total (remember, 3 cigarette butts can be turned into 1 cigarette). He smokes all three of these, and now he has 4 cigarette butts.

He then turns 3 of the 4 cigarette butts into another cigarette and smokes it. He has now smoked 4 cigarettes and has 2 cigarette butts.

For the final step, he goes and borrows one of Tom's cigarette butts. With this cigarette butt plus the 2 he already has, he is able to make his 5th cigarette to smoke. After smoking it, he is left with 1 cigarette butt, which he puts back in Tom's pile so that Tom won't find anything missing.

He then turns 3 of the 4 cigarette butts into another cigarette and smokes it. He has now smoked 4 cigarettes and has 2 cigarette butts.

For the final step, he goes and borrows one of Tom's cigarette butts. With this cigarette butt plus the 2 he already has, he is able to make his 5th cigarette to smoke. After smoking it, he is left with 1 cigarette butt, which he puts back in Tom's pile so that Tom won't find anything missing.

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The flight is full and you are last in line. What is the probability that you get to sit in your assigned seat?

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We first make two observations:

1.

2.

Based on these observations, we know that the instant that a passenger sits in either Steve's seat or your seat, the game for you is "over", and it is fully decided if you will be sitting in your seat or not.

Our final observation is that for each of the first 99 people, it is EQUALLY LIKELY that they will sit in Steve's seat or your seat. For example, consider Steve himself. There is a 1/100 chance that he will sit in his own seat, and a 1/100 chance that he'll sit in your seat. Consider any other person who has been displaced from their own seat and thus must choose a seat at random...if there are N seats left, then there is a 1/N chance that they'll sit in Steve's seat, and a 1/N chance that they'll sit in your seat.

So because there is always an equal chance of a person sitting in your seat or Steve's seat (and one of these situations is guaranteed to happen within the first 99 people), then there is an equal chance that you will or will not get your seat. So the chance you get to sit in your seat is 50%.

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Clockwise – 0

Anti-clockwise – 1

Now there can 8 cases (i.e. 23) which can represent Ants moment as:

There would be only 2 cases when all ants are moving in same direction, in which case they will not collide i.e. Case 1 and Case 8.

So probability will be

2/8 = ¼ = 25%

There are 2 Red and 2 Black hats. Every person is wearing one hat, anyone can tell the color of his hat and all 4 will be free. There is only one chance to say the color by anyone.

Conditions are:

a) Person 1 & person 2 can see only wall.

b) Person 3 can see person 1 and wall.

c) Person 4 can see person 1&2 and wall.

a) If Person 2 & 3 is wearing red caps then person 4 can say his hat color as black without doubt.

b) Similarly, If Person 2 & 3 is wearing black caps then person 4 can say his hat color as red without doubt.

c) Now if Person 2 & 3 is wearing different color caps then Person 4 will not say anything, he will remain silent. For example after 1 min wait person 3 knows that person 4 is not answering that means he and person 2 is wearing different color caps. Then he can see the color of person 2 cap and can tell color of his cap accordingly.

How many shots(maximum shots) you need to make sure that monkey get killed.

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Shoot on Tree 1. Monkey is not there he jump.

Shoot again on Tree 1. If from step 1 he jump to tree 1 then he die. Else shoot twice on tree 2 and so on.

By doing this monkey will be killed or he is moving towards tree 5.

Once you reach at tree 4, monkey will be either at tree5 or tree 4 and he will be killed for sure.

How many shots(maximum shots) you need to make sure that monkey get killed.

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a) Batsman A hit 6 run on first ball.

b) Batsman A again hit ball in air and start running, both batsmen cross over and A was caught by the fielder.

c) Then Batsman B takes the strike and hit the six again.

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In Next ball Batsman B hit a six and makes his century and makes his 100 and wins the match.

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MMM DDD ---

-> MMM D --- DD

-> MMM DD --- D

-> MMM --- DDD

-> MMM D --- DD

-> M D --- MM DD

-> MM DD --- M D

-> DD --- MMM D

-> DDD --- MMM

-> D --- MMM DD

-> DD --- MMM D

-> --- MMM DDD

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a) Ask A, did he die? if he says NO, he’s the one who always lies. Ask B, which is the door to heaven and take that door.

b) If A says Yes, take the other door that B did NOT say went to heaven.

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a) Take a fruit from (box1) orange/apple -> let it be orange.. as it is wrongly labeled it can never be orange/apple. So it is Orange.

b) So the remaining two boxes (box1 and box2) can be either orange/apple or apple.

c) The apple box (say box2) can never be apple itself as it must be wrongly labeled. So it is Orange/apple. So the final box (box3) will be apple.

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a) Take one apple from each box weight 5 on each side.

b) From five divide 2-2-1

c) Then weigh again 2 if required.

a) Take 1 apple from box 1, 2 apples from box 2 ... 10 apples from box 10

b) Put all on scale, say it shows XX.Y pounds

c) Look at Y (fractional part).

d) if it's 0, it's box 10

1 - box 9

2 - box 8

...

9 - box 1

a) Person from TrueValley always say truth.

b) Person from FalseValley always say false.

c) You don’t know the person standing is from which valley.

What will be your question to find your way?

a) Person is from TrueValley then he will take towards his home i.e. TrueValley which is your destination.

b) Person is from FalseValley then he will take towards other route instead of his home which was FalseValley. As he is a lier so will never ever take you towards his home.

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There may be more than one solution, however, I am describing here the one which struck me. We will divide balls in groups of four. Thus we will have three groups, A, B and C. Without loss of generality , put group A and B on comparison balance.

In this case, we know that defective ball is in group C. Put (C1.C2,C3) and ( A1,A2,A3) on comparison balance.

Clearly, C4 is the defective ball. Compare it with A1 and find out whether it is heavier or lighter.

Now, we know that defective ball is lighter. Compare C1 and C2. if they are unequal, we know which one is defective depending upon which one is lighter. If they are equal, we know that C3 is the lighter defective ball.

Now, it is obvious that defective ball is heavier. Carry out the same procedure as described in case 1b, to find out which one is heavier defective ball.

From this observation we can deduce that one of the ball in group A is heavier or one of the ball in group B is lighter. Now, compare ( A4, B3,B4) and ( B1,B2,C1).

This tells us that either A4 is heavier or one of B1-B2 is lighter. Compare B1 and B2. If they are equal then A4 is the heavier defective ball. If not, we know which one is defective depending on which one is lighter.

This tells us that one of B3 and B4 is lighter defective ball.Compare B3 and B4 and find out which one is defective.

This is an easier case. Now we know that one of A1,A2,A3 is defective and heavier. Compare A1 and A2 if they are equal A3 is defective, if not, the heavier of the two is defective.

Line of reasoning is exactly similar to case 2. ( replace heavier by lighter and vice versa ).

When a prisoner is brought out to the lobby, he also has the option of saying "Every other prisoner has been brought out to the lobby." If a prisoner chooses to say this and it is true, all the prisoners will go free. However, if a prisoner chooses to say this and it's wrong, all the prisoners will be executed. So a prisoner should only say this if he knows it is true for sure.

Before the first day of this process begins, all the prisoners are allowed to get together to discuss a strategy to eventually save themselves.

What strategy could they use to ensure their eventual salvation?

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Make one of the prisoners the "lead" prisoner. This prisoner is the ONLY one who is allowed to turn the light off.

Each time any of the other prisoners goes into the lobby, if the light is off, they will turn the light on, but only if they've never turned it on before. This means that each prisoner will only ever turn the light on once.

Meanwhile, every time the lead prisoner goes into the lobby, he will turn the light off if it's on. He will keep track of the number of times he has turned the light off.

Once the lead prisoner turns off the light for the 99th time, he knows that every other prisoner has turned the light on once (and thus has been in the lobby). At this point, he may say that all the prisoners have been to the lobby, and they will all go free.

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Dice 1: 012678

Dice 2: 012345

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