Making it 50-50 in Two!

What was the challenge?

Let us label the 10-liter jug as A, the 7-liters jug as B, and the 3-liters jug as C.

 
STEPS : 

1]  Pour Jug A into Jug C.      A : 7    B : 0    C : 3
 
2]  Pour Jug C into Jug B.      A : 7    B : 3    C : 0

3]  Pour Jug A into Jug C.      A : 4    B : 3    C : 3

4]  Pour Jug C into Jug B.      A : 4    B : 6    C : 0

5]  Pour Jug A into Jug C.      A : 1    B : 6    C : 3

6]  Pour Jug C into Jug B.      A : 1    B : 7    C : 2

7]  Pour Jug B into Jug A.      A : 8    B : 0    C : 2

8]  Pour Jug C into Jug B.      A : 8    B : 2    C : 0

9]  Pour Jug A into Jug C.      A : 5    B : 2    C : 3

10] Pour Jug C into Jug B.     A : 5    B : 5    C : 0

Task Completed.

The Bookworm's Move

There are three books on a shelf standing side by side as they are normally placed on bookshelf. Each book is 10ml thick. The bookworm eats its way through the first page of the first book to the last page of the last book.

How far did the bookworm move if it can only move in a straight horizontal line?

30 mm? It must have moved only 10mm! 

Tricky Data of Bookworm's Movement

What was the data given?

Yes, bookworm only moved 10 mm distance. Read the given data given once again. 

Books are arranged in the way we normally place on the shelf. Normally, they are placed in such a way that the their spines (where most of books have name) are easily visible to us. 

If A, B and C are books having 10 pages each then with the normal placement the pages would count from 10 to 1 if counted from left to right. 

In other arrangement, they would count from 1 to 10 if counted from left to right.

See below.

Now, as per data given, the books are arranged in normal way on the shelf and the bookworm started from first page of first book to the last page of last book.
That is it must have ate only middle book which is 10 mm wide.

Hence, the bookworm must have moved only 10 mm.

The Tunnel Trouble!

A man needs to go through a train tunnel to reach the other side. He starts running through the tunnel in an effort to reach his destination as soon as possible. When he is 1/4th of the way through the tunnel, he hears the train whistle behind him. 

Assuming the tunnel is not big enough for him and the train, he has to get out of the tunnel in order to survive. We know that the following conditions are true - 

1. If he runs back, he will make it out of the tunnel by a whisker.

2. If he continues running forward, he will still make it out through the other end by a whisker.

What is the speed of the train compared to that of the man?

The train must be traveling at THIS speed!

Escape From The Tunnel Trouble!

What was the question?

LOGICAL APPROACH : 

As per condition, if the man runs back he will make it out of the tunnel by a whisker. That means while he runs back 1/4 th tunnel distance, the train travels from it's position to the start of the tunnel. 

In other words, the time taken by man to get back covering 1/4th to the start of the tunnel and the time taken by train to reach at the start of tunnel is same.

So if the man decides to go forward then by time the train reaches at the start of tunnel, man covers another 1/4th tunnel distance i.e. he will be halfway of the tunnel.

At this point of time, the man needs to cover another 1/2th tunnel distance while train has to cover entire tunnel distance. Since, man just manages to escape from accident with train at the exit of tunnel, the train speed has to be double than man's speed as it has to travel distance double of that man travels.

MATHEMATICAL APPROACH : 

Let us suppose - 

M - Speed of Man

T - Speed of Train

D - Tunnel Distance/length

S - Distance between train and the start of tunnel.

As per condition 1, 

Time needed for man to get back at the start of tunnel = Time needed for train to cover distance F to arrive at the start of tunnel

(D/4)/M = S/T  

D/4M =  S/T  .....(1)

As per condition 2,

Time needed for man to move forward at the end of tunnel = Time needed for train to cover distance S + time needed to cover tunnel distance.

(3D/4)/M = S/T + D/T 

Putting S/T = D/4M,

3D/4M - D/4M = D/T

2D/4M = D/T

T/M = 2

T = 2M.

That is speed of the train needs to be double of the speed of the man.

Interestingly, from (1),

D/S = 4M/T

D/S = 2 

D = 2S

S = D/2

That is train is 1/2th tunnel distance away from the start of tunnel. 

Sequel : Story of Distribution of Loot

The five pirates mentioned previously are joined by a sixth, then plunder a ship with only one gold coin.

After venting some of their frustration by killing all on board the ship, they now need to divvy up the one coin. They are so angry, they now value in priority order:

1. Their lives
2. Getting money
3. Seeing other pirates die.

How can the captain save his skin?

This is how he should save himself! 

 The Prequel of the story!

Captain's Life Saving Proposal in Sequel

First read the story of sequel!

Let's name all the pirates as Pirate 6,5,4,3,2,1 as per their seniority. Now, the captain should respond with the logic below to save his skin.

Let's consider the cases where there are different number of pirates left on the ship after getting rid of seniors one by one.

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CASE 1 : 2 Pirates

The captain i.e. Pirate 2 can keep coin with him & obviously vote for himself (1/2 = 50% vote) to approve the proposal.

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CASE 2: 3 Pirates.

The captain i.e. Pirate 3 offers coin to Pirate 1 to get his support (2/3 = 66%) on proposal. Since, Pirate 1 knows what is going to happen if Pirate 3 dies as crew reduced to 2 Pirates as in CASE 1.

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CASE 3: 4 Pirates.

The captain i.e. Pirate 4 offers coin to Pirate 2 thereby getting his support (2/4 =50% votes) to get approval on proposal. Again, here Pirate 2 is smart enough to agree on this proposal as he know what will happen if Pirate 4 is eliminated leaving behind 3 pirates on sheep as in CASE 3.

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CASE 4: 5 Pirates.
Now the captain i.e. Pirate 5 always will be in danger as he can give only coin to only 1 of remaining 4. So he can 'earn' only 1 vote in support of his proposal i.e. only 2/5 = 40% votes. Hence, there is no way his proposal get approval & he should be ready to die.

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CASE 5: 6 Pirates.

Here, Pirate 5 has to agree whatever captain i.e. Pirate 6 offers as if Pirate 6 dies the case reduces to CASE 4 where Pirate 5 can't save himself. However, if both Pirate 6 & Pirate 5 die, then CASE 3 comes into reality where Pirate 2 gets coin. So Pirate 2 would never agree on proposal offered by Pirate 6.

However, if Pirate 6 offers coin to Pirate 4 then he would take it happily as he knows that what will happen of both Pirate 6 & 5 are killed as CASE 3 comes into picture where he has to give coin to Pirate 2 to save himself.

If Pirate 6 offers coin to Pirate 3 as well & earn his support as Pirate 3 also knows what is going to be the case of both 6 & hence 5 get eliminated. The case will be reduced to 4 Pirates as in CASE 3 where Pirate 4 will offer coin to Pirate 2.

The Pirate 6 even choose Pirate 1 to offer coin. Again, Pirate 1 smart to think that what will be the case if 6 & hence 5 are killed. It will be the scenario as in CASE 3 where Pirate 4 offers coin to Pirate 2.

In short, the Pirate 6 will always get support from Pirate 5 always in any case & any one from Pirate 4/3/1 if he offers coin to any one of these three. 

That's how he will get support of 50% (3/6) group to get approval for his proposal thereby saving his own life on approval of proposal offered.

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