Man in a well

Puzzle Level :- Very Easy

PUZZLE

        A man fell into a 30 feet deep well, in one minute he climbs 3 feet up and slips 2 feet down. How much time it will take for him to come out of the well?

ANSWER

        28 minutes

SOLUTION

        Puzzle is very easy and answer looks is 30 minutes. Since he is climbing 3 feet and and falling 2 feet in one minute so he climbs 1 feet per minute. since well is 30 feet deep . It takes  30 minutes to climb.
But this is wrong.

On 28th minute he climbs out of well and won't slip back.

--------------------End--------------------

The Trainee Technician

Puzzle Level :- Very Hard

PUZZLE

           A 120 wire cable has been laid firmly underground between two telephone exchanges located 10km apart. Unfortunately after the cable was laid it was discovered to be the wrong type, the problem is the individual wires are not labelled. There is no visual way of knowing which wire is which and thus connections at either end is not immediately possible. You are a trainee technician and your boss has asked you to identify and label the wires at both ends without ripping it all up. You have no transport and only a battery and light bulb to test continuity. You do have tape and pen for labelling the wires.What is the shortest distance in kilometres you will need to walk to correctly identify and label each wire?

ANSWER

    20km

SOLUTION

        To solve this issue we need to group the wires at one end and go to second end and find the wires in each group with help of battery and light and label them.

Then again form groups by choosing one one wire from previous groups and then go to first end and label them based on their groups.

This is difficult to understand it in statements. so i will explain it in details with example.

For convenience let's assume there are only 6 wires instead of 120. Let's assume them as 1,2,3,4,5,6

Now group them as first group contains one and second group contains two etc...

(Here grouping means we tie the wires and tape them so that they can pass electricity)

{1} , {2,3}, {4,5,6}

and Label each wire with group number (X1,X2,X3 are groups )

{1}  => X1

{2,3} => X2

{4,5,6} => X3

Now go to other end. With battery and light find these groups.

suppose if we attach battery to wire-1 , No other wire will pass the electricity and light will be off

if we pass the current over 2  the current will pass through 2 and 3 similarly for 4,5,6.

This way we can find the group of each wire they belong to

Now label each wire in each group as below

X1 => {A1}

X2 => {A2, B2}

X3 => { A3, B3, C3}

Now group the wires at this second end (Y1,Y2,Y3 are new groups as second end)  by picking one wire from each first group (i.e X1, X2, X3)

{A1, A2, A3} => Y1

{B2, B3}  => Y2

{C3}  =>  Y3

Now go to first end of cable  and separate all wires and try to find the each wire's group(i.e Y1, Y2, Y3) using battery and light.

If the wire belongs to  Y1 group and previously it is in X1 group then label it as A1

similarly wire belong to Y1 and X2  groups is A2

Y1, X3 => A3

Y2, X2 => B2

Y2, X3 => B3

Y3, X3 => C3

This way we can label the wires correctly at both ends.

and we just travelled 20km in this case.

Similarly for 120 wire case we can first group wires as 1,2,3,4,5....etc  and solve the puzzle with similar approach.

--------------------------------END--------------------------

25 Horses puzzle

Puzzle Level :- Moderate

PUZZLE

         You have 25 horses, among these you need to pickup 3 fastest horses for an upcoming competition. Unfortunately you forgotten stop watch at home and there are only 5 tracks to test these horses. (i.e at a time only 5 horses can race)
What is the minimum number of races required to find the 3 fastest horses?

ANSWER

7

SOLUTION

         Since there are 5 tracks we can group these horses into 5 groups, with 5 horses in each group.

and now find the fastest three in each group.

let's number these horses based on fastness in each race. Here A1 is fastest and A5 is slowest in first group and this is same for all other groups.

Race-1 :-  A1  > A2  >   A3  > A4  >  A5  

Race-2 :-  B1  >  B2  >  B3  >  B4  >  B5

Race-3 :-  C1  >  C2  > C3  >  C4  >  C5

Race-4 :-  D1  >  D2  > D3  >  D4  >  D5

Race-5 :-  E1  >  E2  >  E3  >  E4  >  E5

      You might get doubt why we need to pick 3 fastest one in each group, instead of picking the top one from each group and find fastest 3 in these horses (i.e A1,B1,C1,D1,E1).

      This is because second or third fastest one in a group can be faster than top ones from other groups.  i.e for example A2 can be faster than B1,C1,D1,E1 , then we need to pick A2 instead of top one from other groups.

So we can eliminate A4, A5, B4, B5, C4, C5, D4, D5, E4, E5 from all 25 horses

Now conduct Race-6 between fastest one from each group (A1, B1, C1, D1, E1) to find overall fastest one.

Who ever wins this race is the fastest one among all 25 horses

Now let's assume C1 won the race and  the order of finishing race is as below.

Race-6 :- C1 > D1 > B1 > A1 > E1

Keep C1 aside since it is fastest one among all horses.

Since A1 and E1 are slowest in Race-6 we can eliminate these and obviously we can also eliminate A2,A3 and E2, E3.

Since we only need two more fastest horses , we can eliminate B2, B3 because D1,B1 are faster than these two.

And also we can eliminate D3 since D1,D2 are faster than these.

So now remaining horses are

C2, C3, D1, D2, B1

Now conduct Race-7 with these horses (C2, C3, D1, D2, B1).

Which one come in first two places are second and third fastest one among all 25 horses

So in 7 races we can find fastest three horses.

----------------------END--------------------

Defective ball

Puzzle Level :- Moderate

PUZZLE

          There are 9 identical looking metal balls. One of them is defective. You have simple mechanical balance to weigh balls.

What is the minimum number of times we have to use balance to find defective ball?

ANSWER

   4

SOLUTION

                This puzzle is similar to  Heavy ball puzzle. main difference is We don't know the defective ball is heavier or lighter than other balls.  So we can follow the similar approach as in Heavy ball puzzle.

First take 3 , 3 balls in each side of balance and keep 3 balls aside

{1,2,3}(left side) -- {4,5,6}(right side)       {7,8,9}(kept aside)

CASE-1 

        If balance is equal on both sides  then one of {7,8,9} is defective
 then in iteration-2
 keep one one ball on each side of balance and keep one aside
{7}(left side) -- {8}(right side)       {9}(kept aside)

=>if balance are equal , 9 is defective ball (So in this case in two iterations we can defective ball)
=>if balance is unequal, then one of 7,8 is defective but we can't find which is defective
     so in iteration-3  we compare one of 7,8 with any one of good ball
             {7}(left side) -- {9}(right side)
        Now if balance is equal 8 is defective otherwise 7 is defective

CASE-2 

       if balance is unequal then defective ball is in one of these two sets {1,2,3},{4,5,6}
To find which set has defective ball , compare it with good ball set {7,8,9}
{1,2,3}(left side) -- {7,8,9}(right side)
=> if balance is equal then defective ball is in {4,5,6} else defective ball is in {1,2,3}
Now apply same logic explained in case-1 on defective set to find defective ball

So in best case it takes 2 iterations to find defective ball, but in worst case it takes 4 iterations.

This puzzle can be extended to find the defective ball is either heavier or lighter.

----------------------END------------------------

Heavy ball

Puzzle Level :- Moderate

PUZZLE

       There are 8 identical looking metal balls. One of them is heavier than the other 7 balls.
You have simple mechanical balance to weigh balls.

What is the minimum number of times we have to use balance to find heavy ball

ANSWER

    2

SOLUTION

        Generally we think that, by keeping four balls in each side we can eliminate 4 balls 

and then two balls on each side and then one ball on each side to find heavy ball.

but here 3 iterations is needed to find heavier ball.

     We can improve this and find heavier ball in 2 iterations.

I am assigning numbers to each ball for easy explanation 

Total balls {1,2,3,4,5,6,7,8}

First take 3 , 3 balls in each side of balance and keep 2 balls aside

{1,2,3}(left side) -- {4,5,6}(right side)       {7,8}(kept aside)

CASE-1 

        If balance is equal on both sides  then either of {7,8} is heavy then in next iteration we can weigh and find which is heavier by keeping these balls on each side of balance.

CASE-2 

       if balance is heavier on left side{1,2,3} then one of these balls {1,2,3} is heavier

Then in next iteration take one one ball on both sides of balance and  keep one aside

{1}(left side) --{2}(right side)     {3}(kept aside)

if balance is equal then 3 is heavier 

if balance is lean left then 1 is heavier otherwise 2 is heavier

CASE-3 

       if balance is heavier on right side{4,5,6} then one of these balls {4,5,6} is heavier

Then in next iteration take one one ball on both sides of balance and  keep one aside

{4}(left side) --{5}(right side)     {6}(kept aside)

if balance is equal then 6 is heavier 

if balance is lean left then 4 is heavier otherwise 5 is heavier

So in two iterations we can find the heavier ball.

Same logic can be applied even for 9 balls OR  one ball is lighter instead of heavier.

We can follow similar logic for this kind of simple balance problems

----------------------END------------------------

Six pirates and a gold coin

Puzzle Level :- Hard

PUZZLE

           Sixth pirate joined other five pirates mentioned in previous puzzle (http://solve-puzzles.blogspot.in/2016/10/greedy-pirates-and-gold-coins.html) and looted a passenger ship, unfortunately they got only one gold coin from ship

(All rules to mentioned in previous puzzle are applicable here to divide the loot)

They now need to divide 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.

So if given the choice between two outcomes, in which they get the same amount of money, they’d choose the outcome where they get to see more of the other pirates die.

How can you(most experienced pirate) save yourself?

ANSWER

              By giving one coin to either 1st or 2nd or 4th Pirate  (Here you are 6th pirate and 1st pirate is least experienced one)

SOLUTION

             We can follow the similar approach mentioned in previous puzzle (http://solve-puzzles.blogspot.in/2016/10/greedy-pirates-and-gold-coins.html)

Until Case-4 there are no issues , we can follow the same logic.

     In Case-5 , Pirate-5 can't get 3 votes. 

Pirate-3 always votes against since he gets coin anyway in situation as case-4

Besides his vote pirate-5 can get one more vote by giving coin to either pirate-1 or 2 or 4.

He still need one more to save himself.  So in Case-5 always pirate-5 dies.

	Pirate-6	Pirate-5	Pirate-4	Pirate-3	Pirate-2	Pirate-1
Case-2					1	0
Case-3				0	0	1
Case-4			0	1	0	0
Case-5		0	1	0	0	0
Case-6	0	0	1	0	0	0

          In Case-6 ,  Pirate-6 gets his vote  and pirate-5 also votes him because he will die if Case-5 situation arises.  Pirate-3 votes against you even if you give coin to him because, anyway he gets coin if situation moves to case-4 

So You can get another vote by giving coin to either Pirate-1 or 2 or 4

---------------------------------END-------------------------------

Greedy pirates and Gold coins

Puzzle Level - Hard

PUZZLE

       Five pirates looted 100 gold coins by raiding merchant ship and they want to divide these coins among themselves. These pirates are extremely intelligent ,cruel and greedy and each one want major share of booty.

       The pirates are ranked based on their experience. The most experienced pirate gets  chance to propose a plan and then all the pirates vote on it. If at least half of the pirates agree on the plan, the gold is split according to the proposal. If not, the most experienced pirate is thrown off the ship and this process continues with the remaining pirates until a proposal is accepted. 

        If you are the most experienced pirate among them and got first chance to distribute coins, then what is the maximum number of coins you can keep without risking your life?

ANSWER

• Pirate-1(You)  ==> 98 Coins
• Pirate-2           ==> 0 Coins
• Pirate-3           ==> 1 Coin
• Pirate-4           ==> 0 Coins
• Pirate-5           ==> 1 Coin

SOLUTION

       First it looks impossible to get even single coin , because all other pirates can vote against your proposal to get even that single coin from you.

       Even if you not taken single coin also, they might vote against you since each one is such a greedy persons that they want all booty for themselves not giving anything to other pirates.

To solve this problem we need to follow the bottom to top approach.

i.e let's assume there is one pirate then obviously he gets the all the Coins

Now take case of 2-pirates

      First pirate distributes coins as 100-0.  Even second pirate votes against this proposal. proposal is accepted since first pirate got 50% votes since he votes for himself

Let's move on to 3-pirates case

           Apart from your vote, you need another vote to approve your proposal

The next experienced pirate(2nd pirate) will vote against you since he get 100 coins if you die (take look at 2-pirates case)

         You can give 1 gold coin to 3rd pirate and ask to vote in favour, he happily agrees because if he votes against you and move to 2-pirates case he gets no coins.

So you can distribute like 99--0--1 and you can get 99 coins without risking your life

Similarly for case of 4-Pirates

              Apart from you,  you need another vote to accept the proposal.

Next experienced pirate (i.e pirate-2) try to vote against you because he gets 99-coins if you die and situation moves to 3-Pirates case.

But Pirate-3 will favour you because he get 0-coins if situation moves to 3-Pirates case

    You can give 1-coin to 3rd pirate and get his vote.

So distributions is like 99-0-1-0 and you can enjoy 99 coins

Similarly this logic can be extended for 5 ,6,7 ... pirates

In our 5-Pirates case Distribution should be like  98-0-1-0-1. so that you can get maximum booty without risking your life.

---------------------END-----------------------

Mislabeled Baskets

Puzzle Level :-Easy

PUZZLE

              You have 3 baskets of fruits. One basket contains Apple, other contains Oranges and the third basket contains both Apple and Oranges. These baskets are mislabelled and You can't peep into basket to find what it contains.

     You have to label basket correctly. You are allowed to pick as many fruits as you want from each basket to fix the labels on baskets. What is the minimum number of fruits that you have to pick and from which basket to correctly label them?

ANSWER

 Need to pick 1 fruit from basket labelled as Apples&Oranges

SOLUTION

We have three baskets with labels as below

• Basket-1  ==> Apples
• Basket-2  ==> Oranges
• Basket-3  ==> Apples & Oranges

Since here odd one is 3rd basket since it mentions that it has both Apples and Oranges.

Pick one fruit from Basket-3.

It will be either Apple or Orange

Case-1 (Apple is Picked)

 if the Basket-3 contains Apple then Basket-3 should be labelled as Apples

We know all baskets are mislabelled and Basket-2 labelled as Orange

Since it is wrong  the other remaining label is Apples & Oranges

so Basket-2 contains mix of both Apples and Oranges

and obviously Basket-1 will have Oranges

So finally correct labels are

• Basket-1  ==> Oranges
• Basket-2  ==> Apples & Oranges
• Basket-3  ==> Apples

Case-2 (Orange is Picked)

if the Basket-3 contains Oranges then Basket-3 should be labelled as Oranges

Same logic as Case-1 is applicable here, with that logic we can conclude  Basket-1 will have Apples&Oranges

and Basket-2 will have Apples

• Basket-1  ==> Apples & Oranges
• Basket-2  ==> Apples
• Basket-3  ==> Oranges

In both cases if we pick single fruit from Basket-3  which is labelled as Apples & Oranges

we can correct the labels.

--------------END----------

Einstein's Puzzle (Garden puzzle)

Puzzle Level :- Very Hard

Background

           Another flavor of  Einstein's puzzle called Garden Puzzle is solved here 

 it is similar to http://solve-puzzles.blogspot.in/2016/10/einsteins-puzzle.html 

We need similar table approach for this puzzle.

PUZZLE

     Five friends have their gardens next to one another, where they grow three kinds of crops: fruits (apple, pear, nut, cherry), vegetables (carrot, parsley, gourd, onion) and flowers (aster, rose, tulip, lily).

Question:-Who has which garden and what is grown where?

Hints

• They grow 12 different varieties.
• Everybody grows exactly 4 different varieties
• Each variety is at least in one garden.
• Only one variety is in 4 gardens.
• Only in one garden are all 3 kinds of crops.
• Only in one garden are all 4 varieties of one kind of crops.
• Pear is only in the two border gardens.
• Paul's garden is in the middle with no lily.
• Aster grower doesn't grow vegetables.
• Rose growers don't grow parsley.
• Nuts grower has also gourd and parsley.
• In the first garden are apples and cherries.
• Only in two gardens are cherries.
• Sam has onions and cherries.
• Luke grows exactly two kinds of fruit.
• Tulip is only in two gardens.
• Apple is in a single garden.
• Only in one garden next to Zick's is parsley.
• Sam's garden is not on the border.
• Hank grows neither vegetables nor asters.
• Paul has exactly three kinds of vegetable.

ANSWER

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner	Hank	Sam	Paul	Zick	Luke
Variety-1	Apples	Onions	Rose	Aster	Pears
Variety-2	Cherries	Cherries	carrot	Rose	Nuts
Variety-3	Pears	Rose	Gourd	Tulip	Gourd
Variety-4	Rose	Tulip	Onion	Lily	Parsley

SOLUTION

From the puzzle description, we can find below information  
Garden Owners (Paul , Hank ,Sam, Zick, Luke ), 
Fruit Varieties (Apple, Pear, Nut, Cherry), 
Vegetable varieties (Carrot, Parsley, Gourd, Onion) 
Flower Varieties (Aster, Rose, Tulip, Lily).

                We have to follow table approach to solve this puzzle. Choosing right columns and rows is very crucial , without right table we can never solve the puzzle. Always read puzzle carefully and choose table rows&columns appropriately. if we fail here, next steps goes in wrong direction

Since Gardens are in certain order we should take them as Columns.

For rows initially i thought of taking 3 rows for each crop type fruits, flowers and vegetables and fill the table with hints to solve the puzzle (like below), but it is wrong approach to solve puzzle since each garden can have more than one variety from each crop type.  This can be understand after reading hints from puzzle description

Wrong Table

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner
Fruit
Flower
Vegetable

 So based on below hint, i have formulated correct table with crop varieties as rows.

 ==>Everybody grows exactly 4 different varieties

Correct Table

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner
Variety-1
Variety-2
Variety-3
Variety-4

First fill the table from the direct hints
==>In the first garden are apples and cherries.
==>Paul's garden is in the middle with no lily.
==>Pear is only in the two border gardens.

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner			Paul
Variety-1	Apples				Pears
Variety-2	Cherries
Variety-3	Pears
Variety-4

Now there are no direct hints, so try to find the hint which are related to other hints
==>Apple is in a single garden.
==>Only in two gardens are cherries.
==>Sam has onions and cherries.
Apples and Cherries are in Garden-1 
Cherries are in  Sam's garden (Which is not Garden-1 since Sam's garden is not on the border.)
So No more Apples and cherries exists in Other Gardens

Now From hint
==>Luke grows exactly two kinds of fruit.
and also Luke don't own Garden-1 because it has 3 fruits 
so Luke must grow two remaining fruits Pears and Nuts
and from hint
==>Pear is only in the two border gardens.
Luke is in Garden-5 and grows Pears and Nuts
and from  hint
==> Nuts grower has also gourd and parsley.

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner			Paul		Luke
Variety-1	Apples				Pears
Variety-2	Cherries				Nuts
Variety-3	Pears				Gourd
Variety-4					Parsley

From Hint
==>Only in one garden next to Zick's is parsley.
We can conclude Garden-4 belongs to Zick
==>Sam's garden is not on the border.
So Garden-2 belongs to Sam
and first garden belongs to Hank
==>Sam has onions and cherries.

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner	Hank	Sam	Paul	Zick	Luke
Variety-1	Apples	Onions			Pears
Variety-2	Cherries	Cherries			Nuts
Variety-3	Pears				Gourd
Variety-4					Parsley

Now take below hint and try to find garden what satisfies this condition in elimination method.
==>Only in one garden are all 4 varieties of one kind of crops.

Garden-2, 5  not satisfies above rule ,it already have different kind of crop
Garden-3 also not satisfies because of condition
==>Paul has exactly three kinds of vegetable.

Now eliminate one garden from Garden-1 and Garden-4
Garden-1 already have 3 fruits and it can have other fruit variey Nuts also and satisfy the Condition but , If garden has Nuts it will fail at condition
==>Nuts grower has also gourd and parsley.
So Garden-4 has  all same kind of crops  ( crop can be either flowers  or vegetables )

Let's determine it has flowers or vegetables crop
==>Hank grows neither vegetables nor asters.
==>Aster grower doesn't grow vegetables
==> Paul has exactly three kinds of vegetable.
From Above hints we can say Asters must be in Garden-4
So Garden-4 has all Flowers

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner	Hank	Sam	Paul	Zick	Luke
Variety-1	Apples	Onions		Aster	Pears
Variety-2	Cherries	Cherries		Rose	Nuts
Variety-3	Pears			Tulip	Gourd
Variety-4				Lily	Parsley

Now find the variety that will be in 4 gardens  to satisfy the hint
==>Only one variety is in 4 gardens 

Garden-4 and Garden-5 columns are already filled up with no common variety. 
so if one variety from these columns (Aster,Rose.Tulip,Lily,Pears,Nuts,Gourd,Parsley)  fits into other 3 gardens.  then above hint will be satisfied
==>Only in one garden next to Zick's is parsley
==>Hank grows neither vegetables nor asters.
so Gourd,Parsley,Asters are eliminated
==>Pear is only in the two border gardens
so pears eliminated
==>Nuts grower has also gourd and parsley. 
So Nuts also wont fit to have in four gardens
==>Tulip is only in two gardens.
==>Paul's garden is in the middle with no lily.
So Tulip and Lily also eliminated

So Rose is the one that will be in four gardens

If above logic is little complicated to understand, please take each variety and eliminate one by one with help of hints to find variety that will be in 4 gardens.

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner	Hank	Sam	Paul	Zick	Luke
Variety-1	Apples	Onions	Rose	Aster	Pears
Variety-2	Cherries	Cherries		Rose	Nuts
Variety-3	Pears	Rose		Tulip	Gourd
Variety-4	Rose			Lily	Parsley

From hints fill complete table
==>Tulip is only in two gardens.
==>Paul has exactly three kinds of vegetable.
==>Only in one garden next to Zick's is parsley.

	Garden-1	Garden-2	Garden-3	Garden-4	Garden-5
Owner	Hank	Sam	Paul	Zick	Luke
Variety-1	Apples	Onions	Rose	Aster	Pears
Variety-2	Cherries	Cherries	carrot	Rose	Nuts
Variety-3	Pears	Rose	Gourd	Tulip	Gourd
Variety-4	Rose	Tulip	Onion	Lily	Parsley

------------------------------END----------------------

Bridge Cross Puzzle

Puzzle Level :- Easy

             This is one of the category of river crossing puzzles.  There are various name for this problem like  Bridge and Torch problem , Dangerous crossing etc....

PUZZLE

       Four people come to a river in the night. There is a narrow bridge, but it can only hold two people at a time. They have one torch and, because it's night, the torch has to be used when crossing the bridge. Each person walks at different speed. Assume 4 persons are A,B,.C,D.
Person A can cross the bridge in 1 minute,
            B in 2 minutes,
            C in 5 minutes,
     and D in 8 minutes.
When two people cross the bridge together, they must move at the slower person's pace.

The question is, What is the minimum time to cross the bridge?

ANSWER

Minimum time to cross the bridge is 15 Minutes.

SOLUTION

This problem can be extended for more than 4 people  with different crossing times.

So to solve this kind of problems we need to keep some points in mind

•  Always  slowest persons should cross together . and same goes for fastest persons
•  Before slowest persons cross the river make sure there are some fast people on other side to  return torch so Fastest persons should cross in first to return torch.
•  While returning the torch to other side always try to use the fastest person

Initial  positions are 

 A(1),B(2),C(5),D(8) (Torch)=====================       (Time :-0 min)

First send fastest persons i.e A,B so it teams 2 mins

                              C(5),D(8)=====================  A(1),B(2) (Torch)    (Time :-2 mins)

Now torch should return so send A back , because he is fastest person available on other side

         A(1),C(5),D(8) (Torch)=====================    B(2)   (Time :-3 mins)

Now let slowest persons cross the river, It takes 8 mins

                                     A(1)=====================   B(2),C(5),D(8) (Torch)   (Time :-11 mins)

Since B is fastest person on other side , let him return the torch

                  B(2),A(1)(Torch)=====================    C(5),D(8)    (Time :-13 mins)

Now A,B crosses bridge

                  B(2),A(1)(Torch)=====================    C(5),D(8)    (Time :-15 mins)

This kind of crossing puzzle can be easily solved if we follow above 3 points even it became complex with more people

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Einstein's Puzzle

Puzzle Level :- Very Hard

Background

       Albert Einstein as a boy invented this puzzle (though there is no proof of authorship) so it is called Einstein's Puzzle (or Einstein's Riddle).
It is often claimed that only 2% of the world population can solve the puzzle. this claim might be true based on decades ago situation but i think  many people from present generation can solve this puzzle.
like i solved here 

There are various versions of this puzzle using different items / situations ,  but basic idea of puzzle is same

PUZZLE

Situation

1. There are 5 houses in five different colors.
2. In each house lives a person with a different nationality.
3. These five owners drink a certain type of beverage, smoke a certain brand of cigar and keep a certain pet.
4. No owners have the same pet, smoke the same brand of cigar or drink the same beverage.

The question is: Who owns the fish?

Hints

• the Brit lives in the red house
• the Swede keeps dogs as pets
• the Dane drinks tea
• the green house is on the left of the white house
• the green house's owner drinks coffee
• the person who smokes Pall Mall rears birds
• the owner of the yellow house smokes Dunhill
• the man living in the center house drinks milk
• the Norwegian lives in the first house
• the man who smokes blends lives next to the one who keeps cats
• the man who keeps horses lives next to the man who smokes Dunhill
• the owner who smokes BlueMaster drinks beer
• the German smokes Prince
• the Norwegian lives next to the blue house
• the man who smokes blend has a neighbor who drinks water

ANSWER

German owns the fish

SOLUTION

First gather all basic information available in puzzle statement, these details will be useful later for solving the puzzle
Based on above puzzle statements we can find below information
There are 5 houses colored in

• Green
• White
• Yellow
• Red
• Blue

These Houses are occupied by 5 different nationalists.

• Brit
• Swede
• Norwegian
• German
• Dane

Each person has one pet from below list

• Dogs
• Birds
• Cats
• Horses
• Fish

Each person drinks one drink from below list

• Tea
• Coffee
• Milk
• Beer
• Water

Each person smokes one brand of cigarettes from below list

• Pall Mall
• Dunhill
• Blends
• Blue Master
• Prince

This kind puzzles needs table kind of approach to solve them. but here difficult part is to choose which items needs to be considered for rows and which items for columns

Here we need to consider Houses as columns since these houses must be in specific order

Now we will make a table  and fill it with the help of hints

	House-1	House-2	House-3	House-4	House-5
House Color
Nationality
Drinks
Pet
Smokes

First choose the hints those directly helps to fill the table
==>From Hint  "the Norwegian lives in the first house"

==>From Hint  "the Norwegian lives next to the blue house"

==>From Hint  "the man living in the center house drinks milk"

	House-1	House-2	House-3	House-4	House-5
House Color		Blue
Nationality	Norwegian
Drinks			Milk
Pet
Smokes

Now we need to check the hints and find the clues which will fill tables indirectly by some implications.
==>From Hint  "the green house is on the left of the white house"
From this  we know that green and white houses are adjacent
Since Blue house is second one ,   first one cannot be green or white becuase they must be adjacent
So first house can be red or yellow

We know that in first house occupied by Norwegian  and  from Hint "the Brit lives in the red house"
we can Conclude that First house is in Yellow color

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue
Nationality	Norwegian
Drinks			Milk
Pet
Smokes

==>From Hint  "the owner of the yellow house smokes Dunhill"
==>From Hint  "the man who keeps horses lives next to the man who smokes Dunhill"

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue
Nationality	Norwegian
Drinks			Milk
Pet		Horses
Smokes	Dunhill

==>From Hint  "the green house's owner drinks coffee"
From this we can say 3rd house cannot be green since 3rd house owner drinks milk
and from hint "the green house is on the left of the white house"
green house should be at 4th position and white should be on 5th position

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue	Red	Green	White
Nationality	Norwegian
Drinks			Milk	Coffee
Pet		Horses
Smokes	Dunhill

==>From Hint  "the Brit lives in the red house"

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue	Red	Green	White
Nationality	Norwegian		Brit
Drinks			Milk	Coffee
Pet		Horses
Smokes	Dunhill

==>From Hint  "the Dane drinks tea"
Dane can be in 2nd house or 5th house  Other options are not feasible

From Hint  "the owner who smokes BlueMaster drinks beer"
this owner can be in 2nd or 5th house Other options are not feasible

So 2nd and 5th house persons drinks either beer or tea

So in first house owner must drink water

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue	Red	Green	White
Nationality	Norwegian		Brit
Drinks	Water		Milk	Coffee
Pet		Horses
Smokes	Dunhill

==>From Hint  "the man who smokes blend has a neighbor who drinks water "

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue	Red	Green	White
Nationality	Norwegian		Brit
Drinks	Water		Milk	Coffee
Pet		Horses
Smokes	Dunhill	Blend

==>From Hint  "the owner who smokes BlueMaster drinks beer"
This fits only for 5th house

and remaining drink Tea will be in 2nd house
and from Hint "the Dane drinks tea" , Dane is in 2nd house

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue	Red	Green	White
Nationality	Norwegian	Dane	Brit
Drinks	Water	Tea	Milk	Coffee	beer
Pet		Horses
Smokes	Dunhill	Blend			BlueMaster

==>From Hint  "the German smokes Prince"
4th house belongs to German , So final house is for Swede
From Hint "the Swede keeps dogs as pets" , 5ht house has dogs
and from Hint "the German smokes Prince"
So House3 owner smokes remaining brand PallMall

From Hint "the person who smokes Pall Mall rears birds" 3rd house has birds

From Hint "he man who smokes blends lives next to the one who keeps cats" House-1 has cats

	House-1	House-2	House-3	House-4	House-5
House Color	Yellow	Blue	Red	Green	White
Nationality	Norwegian	Dane	Brit	German	Swede
Drinks	Water	Tea	Milk	Coffee	beer
Pet	Cats	Horses	birds	FISH	dogs
Smokes	Dunhill	Blend	PallMall	Prince	BlueMaster

So Now for the Question is  Who owns the fish?

The GERMAN owns it..

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Vases-Flowers

Puzzle Level :- Easy

PUZZLE

Mrs. John has some flowers and vases to decorate hall.

• If she keeps one flower in each vase, one flower will remain
• If she keeps two flowers in each vase, one vase will be empty

How many vases and flowers she has.

ANSWER

3-Vases

4-Flowers

SOLUTION

Let's assume  she has X vases and Y flowers.

so from first statement "If she keeps one flower in each vase, one flower will remain"
we can make equation

X*1 +1= Y       (i,e   each vase * one flower + extra flower = total flowers )

and  from second statement "If she keeps two flowers in each vase, one vase will be empty"

(X-1)*2 = Y    (i,e   (total vases - one extra vase) *   two flowers in each vase  = total flowers )

So equations are
1)  X+1 =Y
2)  (X-1)*2 =Y

replace statement 1 in 2

(X-1)*2 =X+1  ==>    2X-2 = X+1   ==>  2X=X+3

 ==>  2X-X = 3    ==>   X=3

So No of Vases are 3

If we substitute it in first statement

X+1=Y  ==>   3+1 = Y  ==>  Y=4

So no of flowers are 4

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First post

First one

Test1
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Today is 13-Oct-2016. From many years i am hearing about blogging but never gave thought of creating one. At last i started one on today.

I have to educate my self on  this so i will be posting these type of test topic for experiments

Let me try how this first post will visible to others