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.

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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.

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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.

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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.

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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

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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

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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.

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