The River Crossing Challenge!

There are 3 men, two Chimps, and one Gorilla on one side of a river :

• They have a boat but only the men and the Gorilla can row the boat across, so there must always be a human and/or Gorilla on the boat.

• The boat can only carry two people/monkeys.

• If monkeys and humans are together on one side of the river there must be as many or more people than monkeys for the men's safety. 

How can all men and monkeys make it to the other side ? 



Here is the PROCESS by which it can be done! 

Responding to The River Crossing Challenge!

What was the challenge ahead?

Recalling the conditions those need to be followed. 

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• They have a boat but only the men and the Gorilla can row the boat across, so there must always be a human and/or Gorilla on the boat.

• The boat can only carry two people/monkeys.

• If monkeys and humans are together on one side of the river there must be as many or more people than monkeys for the men's safety.

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Here, we go step by step process. (M - Men, G - Gorilla, C - Chimps)

1. The gorilla takes 1 chimp across the river and comes back. 

    (M - 3, G - 1, C - 1 | M - 0, G - 0, C - 1) 

2. Again, gorilla takes 1 man across the river and comes back. 

    (M - 2, G - 1, C - 1 | M - 1, G - 0, C - 1)

Now, here gorilla can't take chimp across the river as that will violate condition 3 on that side. Neither gorilla can take 1 man on other side and return back since number of monkeys on returning side will be more than people again violating condition 3.

3. Next, one man drops gorilla at the other side and bring back chimp.

    (M - 2, G - 0, C - 2 | M - 1, G - 1, C - 0) 

4. Now, 2 men has to cross the river and send back gorilla for the rest of work.

   (M - 0, G - 1, C - 2 | M - 3, G - 0, C - 0) 

5. Finally, gorilla takes 2 chimps across the river in 2 round trips.

   (M - 0, G - 0, C - 0 | M - 3, G - 1, C - 2) 

 

A Round Table Conference

8 people have to be seated at a round banquet table. The seats are number 1-8. 

For this teaser, 5 is opposite 1, 6 is opposite 2 and so on. Likewise, "next to" means one of the neighboring seats only. i.e., 8 is next to 1 and 7.
A and H are the only French speaking people on the table. They need to be seated together. 

B and F should sit opposite each other. 

C, F, and G all know Russian, but don't necessarily need to sit next to each other. In fact, G wants to sit next to someone who knows French. 

D is English and insists on sitting next to at least one other English speaking person, and pat opposite the good looking H. E is the only other English speaker in the group, but he wants to sit next to someone who knows Russian. 

G agrees to sit next to F only on condition that the other side must have a French speaker. C does not sit next to either of them. 

C however, agrees to sit next to B, who is the only Bavarian in the group.
A will not sit next to a Russian or an English speaker.

THIS should be the seating arrangement! 

Seating Arrangement in A Round Table Conference

What was the arrangement challenge?

Given Data : 

8 people have to be seated at a round banquet table. The seats are number 1-8.

1. A and H are the only French speaking people on the table. They need to be seated together. 

2. B and F should sit opposite each other. 

3. C, F, and G all know Russian, but don't necessarily need to sit next to each other. In fact, G wants to sit next to someone who knows French. 

4. D is English and insists on sitting next to at least one other English speaking person, and pat opposite the good looking H. 

5. E is the only other English speaker in the group, but he wants to sit next to someone who knows Russian. 

6. G agrees to sit next to F only on condition that the other side must have a French speaker. C does not sit next to either of them. 

7. C however, agrees to sit next to B, who is the only Bavarian in the group.
A will not sit next to a Russian or an English speaker.

Arranging STEPS :

1] From (1) & (3), it's clear that G should be seating next to the A or H. That is G, A and H are occupying adjacent seats whose order is yet to be known. 

2] Now as per (2), B and F must be on opposite seats, they need to 'surround' the group of G - A - H. That's the only way they would be opposite. (There are 2 ways that they can do this and this would result into 2 possible solutions as concluded in conclusion).

3]  As per (6), G is ready to sit next to F so that G and F should be occupying adjacent seats. Order of seat to be occupied by A and H are yet to be known.

4] As per (7), C can seat next to B. A can't seat next to Russian G. 

5] As per (4) and (5), D & E should be next to each other but as per (4) D has to be opposite to H. 

CONCLUSION : 

In order to make seat arrangement in accordance with everyone's preference, we just need to ask them to seat in alphabetical order of their names. Here, seat number doesn't matter and what matters is that everyone is getting their favorite seat.

Interestingly, they can seat in circle clockwise or anticlockwise according to alphabetical order of their names.

 

Confusing Ride on the Ferris Wheel

There are 10 two-seater cars attached to a Ferris wheel. The Ferris wheel turns so that one car rotates through the exit platform every minute. 

The wheel began operation at 10 in the morning and shut down 30 minutes later. 

What's the maximum number of people that could have taken a ride on the wheel in that time period?

Here is count of people enjoying the ride!

Peoples Enjoying Ferris Wheel Ride

What was the puzzle?

Let's simplify the situation by naming 10 cars as A, B, C, D, E, F, G, H, I, J.

Suppose the Car A is at the exit platform at 10:00 AM. Obviously it can be 'loaded' with 2 peoples say A1-A2.

At 10:01 AM, the Car B will be at the exit platform which can be 'loaded' with 2 more peoples say B1-B2.

So continuing this way, at 10:09 AM the car J will be loaded with peoples J1-J2.

At 10:10 AM, the Car A will be again at the exit platform and now A1-A2 can get off the board to allow 2 more new peoples A3-A4 to get on the board.

At 10:11 AM, the Car B will be at the exit platform where B3-B4 will replace B1-B2. 

Continuing this way, at 10:19 AM the J1-J2 in the Car J will be replaced by J3-J4.

So far, 10 x 4 = 40 different peoples would have enjoyed the ride. 

It's clear that every car takes 10 minutes to be at the exit platform after once it goes through it. That's how Car A is at the exit platform at 10:10 AM, 10:20 AM and it can be at 10:30 AM as well when the wheel is supposed to be shut down.

Now, since wheel needs to be shut down 10:30 AM, emptying all the cars must be started except Car A for the reason stated above.

Therefore, at 10:20, peoples A5-A6 replace A3-A4. Thereafter, every car should be emptied. So, far 40 + 2 = 42 different people have boarded on the cars of the wheel.

So, at 10:21 AM, the Car B is emptied, at 10:22 AM, the Car C should be emptied and so on.

At 10:29 AM, J3-J4 of Car J get out of the car and finally at 10:30 AM, A5-A6 get out of the Car A. 

Now, the ferris wheel can be shut down with no one stuck at any of cars.

To conclude, 42 different peoples can enjoy the ride in given time frame.

What Was the Color of Gabbar Singh's Shirt ?

In Rangeelia, a neighbour situated west of our country, the native people can be divided into three types: Lalpilas, Pilharas and Haralals. 

Lalpilas always get confused between red and yellow (i.e. they see yellow as red and vice versa) but can see any other color properly. Pilharas always get confused between yellow and green (i.e. they see yellow as green and vice versa) but can see any other color properly. Haralals always get confused between green and red (i.e. they see green as red and vice versa) but can see any other color properly.

Three people Amar, Akbar and Anthony, who belong to Rangeelia, made the following statements about Gabbar Singh, the famous dacoit of Rangeelia, when he was last seen by them :-

Amar : Gabbar Singh was wearing a green shirt.

Akbar : Gabbar Singh was not wearing a yellow shirt.

Anthony : Gabbar Singh was wearing a red shirt.

If none of Amar, Akbar or Anthony is a Haralal, then what was the color of Gabbar Singh's shirt ?

THIS is the color of Gabbar Singh's shirt! 

Plan The Best Chance of Winning!

You are playing a game of dodge ball with two other people, John and Tom. You're standing in a triangle and you all take turns throwing at one of the others of your choosing until there is only one person remaining. You have a 30 percent chance of hitting someone you aim at, John has a 50 percent chance, and Tom a 100 percent change (he never misses). If you hit somebody they are out and no longer get a turn.

If the order of throwing is you, John, then Tom; what should you do to have the best chance of winning?

This should be you plan to increase chances of winning! 

Planning The Best Chance of Winning!

What was the game?

You should miss the first shot for the purpose.

Remember, about one of your 3 shots (30% accuracy), John's 1 out of 2 shots (50% accuracy) and Tom's every shot is on target.

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CASE 1 : If you target Tom and hit him then John has to hit you. Even if he fails to target you in first attempt he will be successful in his second attempt. 

And since, your first shot was on target your next 2 has to be off the target one of which will give John a second chance.


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CASE 2 : If you target John then Tom will certainly hit you to be winner of the game.

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CASE 3 : Better miss the first shot and then 1 of next 2 shots will be on the target.

Now, John has to target Tom otherwise assuming John as stronger player, Tom will eliminate him immediately. 

    CASE 3.1 : If John hits Tom and eliminates him then it's your turn

                     now and John's next attempt has to be off the target. 

                     So, even if you fail in first try after John's

                      unsuccessful try you can surely hit John in second try.

    CASE 3.2 : And if John misses Tom then Tom will throw John 

                     out of game in his first attempt. Now, it's your turn 

                     and you can target Tom with 50% accuracy.

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This is the best plan to get a chance of winning this game. 

Language Barrier in International Meeting

Of the 1985 people attending an international meeting, no one speaks more than five languages, and in any subset of three attendees, at least two speak a common language. Prove that some language is spoken by at least 200 of the attendees.

Here is the proof! 

For The Communication in International Meeting

What was the barrier in the communication?

For any attendee A and B, having no common language there must be C who know the language of either A or B to form a trio as mentioned.

Let's make assumption contradicting the statement made in question. Suppose there are only 198 people who can talk in particular language with A or B. Since A can communicate in 5 languages, there are 5 x 198 = 990 people who can talk with A.

That is 990 people are there who have sharing 1 common language with 1 of 5 languages known by A. Similarly, B also can communicate with 990 more people.

Now, if A and B have no common language then there are only 990 + 990 = 1980 people having potential to become C in the trio. This obviously doesn't cover total of remaining people i.e. 1985 - 2 (A and B) = 1983.

Hence, our assumption goes wrong there. So there must be at least 200 attendees knowing the same language .

Interesting Fact of Handshake Count

Suppose we fill Yankee Stadium with 50,000 people and ask them to spend the day shaking hands with one another.

Prove that, at the end of the day, at least two participants will have shaken hands with the same number of people.

Click here for proof!

Proving Interesting Fact of Handshake Count

What was that fact?

Let's contradict the given fact & assume no 2 people have same number of handshakes. In that case, the most gregarious person would have 49999 handshakes & next gregarious person would shake hands with 49998 people and so on. 

This way, the shyest person should have 0 shake hands. But the most gregarious guy must have had handshake with this shyest guy as his count of 49999 also includes this shyest guy. So this is impossible case.

Hence, at least 2 participants would have shaken hands with the same number of people.

 
In other way, the most shyest participant would have 1 handshake, next shyer guy would have 2 & so on. The more gregarious would have 49999 handshakes that includes the shake hand with the most gregarious person. Now, most gregarious person is bound to have 49999 handshakes as he/she can't have 50000 as there are only 50000 people in the stadium including himself/herself. 

That's why, at least 2 participants would have shaken hands with the same number of people.

Truth Tellers and Liars in Circle

On a certain island there live only knights, who always tell the truth, and knaves, who always lie.

One day you find 10 island natives standing in a circle. Each one states: "Both people next to me are knaves!"

Of the 10 in the circle, what is the minimum possible number of knights?

Do you think there can be 5?

Identifying Number of Truth Tellers in Circle

What was the task given?

Recalling the given situation. 

On a certain island there live only knights, who always tell the truth, and knaves, who always lie.

One day you find 10 island natives standing in a circle. Each one states: "Both people next to me are knaves!"

 
Every Knight must be surrounded by 2 Knaves and every Knave has to be surrounded by at least one knight to satisfy the given condition. So there must be Knave-Knight-Knave groups must be standing in circle. 

Now if Knave of previous group is counted for the next group, then there will be 5 knights in the circle as shown below.

But the question asks minimum possible number of Knights in the circle.

So after forming 3 groups of Knave-Knight-Knave separately (total 9 in circle), the last person will be obviously surrounded by 2 knaves. Hence, he must be Knight. See below.

This way there can be only 4 knights standing in the circle without violating the given condition.
 

Count The Number of People From Handshakes

At a party, everyone shook hands with everybody else. There were 66 handshakes.
How many people were at the party?

Skip to the count!

Getting Count of Number of Peoples From Handshakes

Want to read question first? Click here!

Let's suppose that there are 'n' number of people in the party.

The first person will shake hand with (n-1) people, the second person will shake hand with (n-2) people, the third will shake hand with (n-3) people.

In this way, (n-1) th person will shake hand with n-(n-1) = 1 person i.e. last person.

Adding all the number of handshakes,

(n-1) + (n-2) + (n-3) + ..... + 3 + 2 + 1 = n[(n-1)/2]

But total handshakes given are - 66

n[(n-1)/2] = 66

n(n-1) = 132

n^2 - n - 132 = 0

(n-12)(n+11) = 0

n = 12 or n = -11

Since number of people can't be negative, n = 12.

Hence there are 12 people in the party.

Case of 3 Identical Notebooks

Three people all set down their identical notebooks on a table. On the way out, they each randomly pick up one of the notebooks. What is the probability that none of the three people pick up the notebook that they started with?

That's correct probability!

Probability in Case of 3 Identical Notebooks

What was the case?

Let's name peoples as Person - 1, Person - 2, Person - 3 and their notebooks as Notebook - 1, Notebook - 2 and Notebook 3 respectively.

Now there can be 6 ways 3 notebooks can be distributed among 3 persons like below.

(Here, for convenience, 3 different colors are assigned to the notebooks of 3 persons.)

As we can see, there are only 2 cases, where the each of person not getting own notebook. In rest of cases, at least 1 person got own notebook & notebooks of others are shuffled between 2.

So the probability that none of the three people pick up the notebook that they started with is 2/6 = 1/3. 

Lighting Up The Candles

In a group of 200 people, everybody has a non burning candle. On person has a match at lights at some moment his candle. With this candle he walks to somebody else and lights a new candle. Then everybody with a burning candle will look for somebody without a burning candle, and if found they will light it. This will continue until all candles are lit. Suppose that from the moment a candle is lit it takes exactly 30 seconds to find a person with a non burning candle and light that candle.

From the moment the first candle is lit, how long does it take before all candles are lit?

ESCAPE to answer!