The Water Jug Challenge : Puzzle

You've got a 4 liter jug and a 9 liter jug. 

You've got a pool of water. 

What is the fewest number of steps it takes to come up with exactly 6 liters of water? 

No, you cannot pour some water into the 9 liter jug and then guess. Nor can you fill each jug up half way or something. You have to be exact. A step is defined as putting water into a jug (not emptying). For instance, filling the 4 liter jug and then pouring it into the 9 liter jug is 2 steps.

Click here for SOLUTION! 

The Water Jug Challenge Puzzle : Solution

What was the puzzle?

We've got a 4 liter jug and a 9 liter jug. 

We've got a pool of water. 

We have to find fewest number of steps it takes to come up with exactly 6 liters of water. 

STEPS : 

1. Fill up the 9 liter jug from the pool.         9L - 9  4L - 0

2. Fill up the 4 liter jug from the 9 liter jug. 9L - 5  4L - 4

3. Empty the 4 liter jug into the pool.          9L - 5  4L - 0

4. Fill 4 liter jug again from the 9 liter jug.   9L - 1  4L - 4

5. Empty the 4 liter jug into the pool           9L - 1  4L - 0

6. Transfer 1 liter left 9L jug into the 4L.     9L - 0  4L - 1

7. Fill the 9 liter jug from the pool.             9L - 9  4L - 1

8. Fill up the 4 liter jug from the 9 liter jug. 9L - 6  4L - 4

That's how we have exactly 6 liters of water in 9 liters jug. 

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! 

Aliens Meeting on the Earth

100 aliens attended an intergalactic meeting on earth.

73 had two heads,
28 had three eyes,
21 had four arms,
12 had two heads and three eyes,
9 had three eyes and four arms,
8 had two heads and four arms,
3 had all three unusual features.

How many aliens had none of these unusual features?

These are Aliens not having unusual features! 

Normal Aliens in the Meeting on the Earth

Read given data first!

Let's simplify the process with Venn diagram like below.

1. We know, 3 had all three unusual features.

2. 12 had two heads and three eyes, 9 had three eyes and four arms, 8 had two heads and four arms. Here, 3 having all unusual features too need to be counted in each of Aliens' counts above. 

So, Aliens having 2 unusual features of 2 heads & 3 eyes = 12 - 3 = 9.

Aliens having 2 unusual features of 3 eyes & 4 arms = 9 - 3 = 6.

Aliens having 2 unusual features of 2 heads & 4 arms = 8 - 3 = 5.

3. We know, 73 had two heads, 28 had three eyes, 21 had four arms.

Therefore, the number of Aliens having only 2 heads as an only unusual feature is 73 - (9 + 3 + 5) = 56.  

The number of Aliens having 3 eyes as an only unusual feature is 
28 - (9 + 3 + 6) = 10.

The number of Aliens having 4 arms as an only unusual feature is 
21 - (5 + 3 + 6) = 7.

4. So, we have 56 + 9 + 10 + 3 + 5 + 6 + 7 = 96 Aliens having at least one unusual feature. 

Therefore, the number of Aliens not having any of unusual features is 
100 - 96 = 4.

Crack And Win $500,000 - Puzzle

You are in a game show with four other contestants. The objective is to crack the combination of the safe using the clues, and the first person to do so will win $500,000.

The safe combination looks like this:

??-??-??-??

A digit can be used more than once in the code, and there are no leading zeroes.

Here are the clues:

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.

And get moving, I think another contestant has almost figured it out!

Click here for the SOLUTION! 

Crack And Win $500,000 Puzzle - Solution

What was the puzzle?

We know, the safe has a lock having 4 sets of 2 digits as -

?? - ?? - ?? - ??

Since, leading zeros are not allowed any of the set can't be started with 0 like 01, 07, 09 etc. 

Take a look at the clues given - 

-------------------------------------------------------

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.


------------------------------------------------------- 

STEPS :

1] As per clue (2), digits 2, 3, 4 or 5 aren't allowed in any set. That means only digits 0, 1, 6, 7, 8, 9 are allowed for sure.

2] For (3) to be true, possible combinations of 3rd and 4th sets are - 

10 x 4 = 40
11 x 4 = 44
16 x 4 = 64
17 x 4 = 68
18 x 4 = 72
19 x 4 = 76

The third set can't be 10 for 2 reasons. - 1} As per clue (1), the first set will be 01 and leading 0's are not allowed. 2} The 4th set will have digit 4 which is not allowed.

The third set can't start with 2X, 3X, 4X or 5X as those digits aren't allowed. Moreover, it can't be started with 6X, 7X, 8X as in that case the value of the 4th set will exceed it's maximum possible value of 96 (if digit 2 was allowed) or 76.

Out of all above combinations, only 17 x 4 = 68 and 19 x 4 = 76 are valid combination as rest of combinations have digits that aren't allowed.

So, one thing is sure that the first digit of the third set is 1. And hence the first digit of first set also must be 1 as per clue (1).

3] As per clue (5), the second set can't exceed 20. It can't start with 0. Hence, possible values ranges from 10 to 19. That means, the first digit of the second set is also 1.

4] Now as per clue (4), if you add the first number in the first set with the first number in the second set you will get 8. That means the first number of first set is 8 - 1 = 7. 

As of now, the code looks like : 71 - 1? - 1? - ??

5] If 7 is the first digit of first set then as per clue (1), 7 itself is second digit of the third set.

Now, the code looks as : 71 - 1? - 17 - ??

6] As per clue (3) and rightly deduced as a possible combinations for 3rd & 4th set in STEP 2, the 4th set must be 17 x 4 = 68.

With that, code turns into : 71 - 1? - 17 - 68.

7] Finally, as per clue (6), the second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set. This multiplication i.e. ? x 7 must be equal to 71 + 1 = 72. 
Hence, ? = 9.

The final code looks like : 71 - 19 - 17 - 68.

  

"How Did They Cross The River?"

Five men and five dogs (each man owned a dog) went hiking. They encountered a river that was swift and deep. The only way to cross it was an abandoned boat, left ashore on their side. But it would only hold three living things. 

Unfortunately, the dogs were edgy and could not be near another person (not even momentarily) unless its owner was present. One of the dogs attended a highly advanced, highly specialized obedience school and therefore knew how to operate the boat (as the men did) -- the other dogs lack this skill. 

How did the five men and the five dogs cross the river?

THIS is how they accepted the challenge! 

The Challenge of River Crossing

What was the challenge?

Let's name five men as M1, M2, M3, M4, M5 & five dogs as D1, D2, D3, D4, D5.
Assume D1 be the dog having advanced skills of operating boat.

TRAVEL CHART :

1] The dog D1 rows D2, D3 across the river and returns back. After returning back D1 takes D4 across the river & returns back once again.

START : D1-M1, D5-M5, M2, M3, M4  DESTINATION : D2, D3, D4

2] Now, M2, M3 and M4 cross river.

START : D1-M1, D1-M5  DESTINATION : D2-M2, D3-M3, D4-M4

3] Someone from DESTINATION needs to return back to allow others at START to cross the river. So, D4-M4 returns and D1-M1 cross the river. 

START : D4-M4, D5-M5  DESTINATION : D1-M1, D2-M2, D3-M3

4] After D1-M1 reaches to the DESTINATION, D3-M3 returns back.

START : D3-M3, D4-M4, D5-M5  DESTINATION : D1-M1, D2-M2

5] Now, M3, M4 and M5 cross the river.

START : D3, D4, D5  DESTINATION : D1-M1, D2-M2, M3, M4, M5.

6] Finally, the dog D1 at the DESTINATION returns back and makes 2 trips to take D3, D4 and D5 across the river.

START : None  DESTINATION : D1-M1, D2-M2, D3-M3, D4-M4, D5-M5.

This way, five men and five dog cross the river successfully.

The Discounted Purchases : Puzzle

Until last week, Mandy had been putting off the purchase of four major items for her home, waiting for her town's annual bargain days. Each year, the town's four main stores (including Nickels) each send out a coupon for a different percentage discount (10%, 15%, 20%, or 30%) on the purchase of a single item. Can you find the discount that Mandy received on each of her buys, and determine from which store she purchased each item?

Given:

Discounts      : 10%, 15%, 20%, 30%
Store Names : Stears, Pal-Mart, Nickels, Dullard's
Appliances    : Computer, Dishwasher, Lawnmower, VCR

Clues:

1) The percentage discount Mandy received on her new computer was twice that of the item she bought at Dullard's. 

2) The discount on the Stears purchase was ten percentage points lower than the one on the riding lawnmower. 

3) The discount on the dishwasher wasn't 15%. 

4) Pal-Mart, (which didn't issue the 30% off coupon) isn't the store at which Mandy purchased the VCR.

Click here for SOLUTION!

The Discounted Purchases Puzzle : Solution

What was the challenge?

First let's have a given data.

Given:

Discounts: 10%, 15%, 20%, 30%
Store Names: Stears, Pal-Mart, Nickels, Dullard's
Appliances: Computer, Dishwasher, Lawnmower, VCR

Clues:

1) The percentage discount Mandy received on her new computer was twice that of the item she bought at Dullard's. 

2) The discount on the Stears purchase was ten percentage points lower than the one on the riding lawnmower. 

3) The discount on the dishwasher wasn't 15%. 

4) Pal-Mart, (which didn't issue the 30% off coupon) isn't the store at which Mandy purchased the VCR. 

 STEPS : 

1] As per clue (1), Dullard's didn't sell computer to Mandy and she didn't receive 10% or 15% off on computer. Also, Dullard's didn't offer her 20% or 30%.

2] The clue (2) suggests that the lawnmower wasn't purchased at 10% or 15% discount. Also, Stears discount must be 10% or 20%. And Stear didn't sell lawnmower for sure.

3] The clue (3) suggests that dishwasher wasn't sold at 15% discount. So, it's clear that Stears must have sold dishwasher at 10%/20% discount and Dullard's must have sold VCR at 15% discount. Therefore, computer must had been bought with 30% discount.

4] And hence, lawnmower at 20%. With that, Stears must have sold dishwasher at 10% discount as per clue (3).

5] Finally, as per clue (4), Pal-Mart didn't offer 30% discount suggests that it must not have sold computer. Therefore, Nickels sold computer at 30% discount and Pal-Mart sold lawnmower at 20% discount.

CONCLUSION : 

Final summary of the all purchases in form of table looks as - 

Make it 50-50 in Two!

You have three jugs:

A 10-liter jug, filled with water
A 7-liter jug, empty
A 3-liter jug, empty

Your objective is to end up having 5 liters of water in the 10-liter jug, and 5 liters of water in the 7-liter jug.

Note that you have nothing else at your disposal other than these three jugs and that you cannot perform measurements by eye or based on the shape of the jugs.

THIS is how it can be done! 

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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Story of Distiribution of 100 Coins Loot

Five ship’s pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).

The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.

If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.

What is the maximum number of coins the captain can keep without risking his life?

He can take away 98 coins! How? Read here!