RIDDLES -- Dimag ki batti jalao

htomar said:
Five 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?
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Hmmm...Similar question was posted long before :-
http://www.jatland.com/forums/showthread.php?p=76287#post76287

[To summarize, the captain can have 98 gold coins]

Pasting the solution I gave :-
Lets number the persons from N1 to N5----N1 being the juniormost and N5 being the seniormost....
In the riddle,its given that these people are very very greedy and logical too......
So what the N5 can do.....Here is my answer.........
Lets apply induction.

Case1: only one person.
If there is only one person,then he will have all the share.

Case2: two persons
let it be N1 and N2.In this case N2 would want to take all the coins,and he has 50% votes (his own vote) for
doing this. This way N1 won't be getting anything.

Case3: three persons
let it be N1,N2, and N3.So in this case N3 needs support of one more member.N2 wont support him,as he knows that if he is able to get rid of N3,then only two persons will be left and he will be at gain(see case2).So N3 would lure N1 for 1 gold coin.since each of them is extremely greedy and logical,so N1 would think that 1 coin is better than nothing,so he would accept his offer.In this way,N3 will have 99 coins,N2 will have none and N1 will have 1.

Case4: four persons
Let it be N4,N3,N2,and N1. then N4 needs support of at least one member. But he wont get it from N3 (from case3) and recursively from N1. So the only person left is N2. Again, N4 will have 99coins,and N2 will have 1.

Case5: five persons
N5 needs at least 2 more people on his side,to get his plan proposed......If we think logically,like them, we know that N4 would want to kill N5,since he is the next seniormost.......So he will lure N3 with one coin.N3 knows that if N4 comes,he won't get anything,so he accepts that deal...Next comes N2...Now he also would want to kill N5,since he would get advantage then...So the one that will be left is N1.The N5 person can share 1 more coin with N1.

So the solution scheme can be:
Person# No. of coins
N5 98
N4 0
N3 1
N2 0
N1 1

Similarly, there can be more solutions,(if N5 is less greedy),then
Person# No. of coins OR
N5 97 97
N4 0 0
N3 2 1
N2 0 0
N1 1 2
 
You have an old-fashioned refrigerator with a small freezer compartment capable of holding seven ice cube trays stacked vertically. But there are no shelves to separate the trays, and if you stack one tray on top of another before the ice cubes in the bottom tray are fully frozen, the top tray will nestle into it, and you won't get full cubes in the bottom tray. You have an unlimited supply of trays, each of which can make a dozen cubes.
What's the fastest way to make full-sized ice cubes?
 
answer of previous post....

You can make 120 cubes (10 full trays) in the time it takes to freeze two trays. First, fill four of the trays with water and turn the other three upside down and use them to space the four apart. That gives you 48 cubes. Next, empty the four trays and put two ice cubes in diagonally opposed corners of each of six of the trays. Fill the remaining holes -- and the entire seventh tray -- with water. Using the ice cubes to hold the trays apart, stack all seven (the seventh tray should go on top), and freeze them. You'll get an additional 72 cubes. You can get 72 cubes for every batch except the first, for which spacer ice cubes are not yet available
 
three young girls Jyoti, Renu and Suman are telling about their ages.

Suman: The difference between Renu's age and mine is three years.
Jyoti: Renu is the youngest.
Renu: Either I am 24 years old or Jyoti 25or Suman 26.
Jyoti: All are above 24 years of age.
Suman: I am the eldest if and only if Renu is the youngest.
Renu: Suman is elder to me.
Jyoti: I am the eldest.
Renu: Suman is not 27 years old.
Suman: The sum of my age and Jyoti's is two more than twice Renu's age.

They also conveyed that one of them had been telling a lie throughout whereas others had spoken the truth and nothing but the truth.
Can you figure out how old is each?
 
Conclusion from the statements....
Suman had been telling a lie throughout.
Renu's age is 25
Suman's age is 26
Jyoti's age is 27or greater than 27........
sanjeev_balyan said:
three young girls Jyoti, Renu and Suman are telling about their ages.

Suman: The difference between Renu's age and mine is three years.
Jyoti: Renu is the youngest.
Renu: Either I am 24 years old or Jyoti 25or Suman 26.
Jyoti: All are above 24 years of age.
Suman: I am the eldest if and only if Renu is the youngest.
Renu: Suman is elder to me.
Jyoti: I am the eldest.
Renu: Suman is not 27 years old.
Suman: The sum of my age and Jyoti's is two more than twice Renu's age.

They also conveyed that one of them had been telling a lie throughout whereas others had spoken the truth and nothing but the truth.
Can you figure out how old is each?
Click to expand...
 
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htomar said:
Conclusion from the statements....
Suman had been telling a lie throughout.
Renu's age is 25
Suman's age is 26
Jyoti's age is 27or greater than 27........
Click to expand...

good analysis tomar,
in Jyoti's age, u didnt notice Renu's statement, she is saying Suman is not 27 and u know, she always tells the truth, that means ofcourse Suman is not 27 but somebody else is 27, than only Jyti can be 27, anyway good cracking
 
Sanjeev ji I notice that statement..she is saying Suman is not 27.
But that doesn't mean somebody else is 27..It may be or not...
sanjeev_balyan said:
good analysis tomar,
in Jyoti's age, u didnt notice Renu's statement, she is saying Suman is not 27 and u know, she always tells the truth, that means ofcourse Suman is not 27 but somebody else is 27, than only Jyti can be 27, anyway good cracking
Click to expand...
 
There are eight pills of identical shape, size and color. 7 pills have same weight as well and are of medicinal value to Tai Rambhateri, 8'th one differs very slightly in weight and is poisonous. Only way to tell the difference is compare it with one of other 7 pills on a balance. You are allowed to use the balance only twice.
Save Tai Rambhahteri's life by identifying the poisonous pill among them.
 
vikrantsiwag said:
There are eight pills of identical shape, size and color. 7 pills have same weight as well and are of medicinal value to Tai Rambhateri, 8'th one differs very slightly in weight and is poisonous. Only way to tell the difference is compare it with one of other 7 pills on a balance. You are allowed to use the balance only twice.
Save Tai Rambhahteri's life by identifying the poisonous pill among them.
Click to expand...

Make 3 sets, two containing 3 pills each, and the third one having 2 pills. Now let the names of the sets be A, B and C.

(i) Weigh A and B sets. If they are equal:-
(a) Discard A and B. Compare the pills in set C to find the poisonous
one.
(ii) If weight (A) > weight (B):-
(a) Now take any two pills from the set B and compare
them. If they are also equal, then the poisonous pill is the remaining
one in set B; else it is one of the two that were weighed.
(iii) If weight (B) > weight (A):-
Follow the procedure (ii)(a) , with the pills of set A to find the
poisonous pill.
 
Neetu ji .u r right only if we know that particular pill is heaviear or lighter...but it is not given in this puzzle....
neetu21 said:
Make 3 sets, two containing 3 pills each, and the third one having 2 pills. Now let the names of the sets be A, B and C.

(i) Weigh A and B sets. If they are equal:-
(a) Discard A and B. Compare the pills in set C to find the poisonous
one.
(ii) If weight (A) > weight (B):-
(a) Now take any two pills from the set B and compare
them. If they are also equal, then the poisonous pill is the remaining
one in set B; else it is one of the two that were weighed.
(iii) If weight (B) > weight (A):-
Follow the procedure (ii)(a) , with the pills of set A to find the
poisonous pill.
Click to expand...
 
htomar said:
Neetu ji .u r right only if we know that particular pill is heaviear or lighter...but it is not given in this puzzle....
Click to expand...

neetu21 said:
Harendra ji, it's given that they differ slightly in weight :)!
Click to expand...

Well, Harendra ji is right to point out that it can never be solved with spec "slight difference" in weight, the poisonous pill must be specified being heavier or lighter in weight from other pills. Otherwise this can't be solved in infinite tries even. I twitched the question expecting the same result as posted by Neetu ji, as she didn't read the question carefully and posted pre-known answer.

If Neetu ji's solution is considered in sync with the specs in the question i posted the probability that Tai Rambhateri won't survive is still 50%.:p:p:D

For the sake of correct question, consider the poisonous pill being little heavier than others. Now ur answer is correct Neetu ji.

Kudos Harendra ji for catching that.
 
I read that "slight difference" weight, then only posted the answer:p, otherwise as Harendra said, the question could not be solved...So I read the question very well Vikrant.

vikrantsiwag said:
Well, Harendra ji is right to point out that it can never be solved with spec "slight difference" in weight, the poisonous pill must be specified being heavier or lighter in weight from other pills. Otherwise this can't be solved in infinite tries even. I twitched the question expecting the same result as posted by Neetu ji, as she didn't read the question carefully and posted pre-known answer.

If Neetu ji's solution is considered in sync with the specs in the question i posted the probability that Tai Rambhateri won't survive is still 50%.:p:p:D

For the sake of correct question, consider the poisonous pill being little heavier than others. Now ur answer is correct Neetu ji.

Kudos Harendra ji for catching that.
Click to expand...
 
vivektaliyan said:
Are bhai is ka ans to bata de....:o
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1) Fill 5 lts vala can.
2) Usme se 3 lts vala bhar lo.
3) Ab 3 lts vala khali kar do, 5 lts vaale mein jo 2 lts hai, usko 3 lts vaale mein bhar do.
4) Fill the 5 lts vala can again. Ab isme se 3 lts vala bharo, then you will have 4 lts remaining in it.

Five Three
5 0
2 3 [5 => 3]
0 2
5 2
4 3
 
Neetu mai to nahi solve kar paya abhi tak 2 tries mai..aapne kiya hai to answer post kar do...
if we dont know whether that pill is lighter or heavier then it takes 3 attempts..
may u have a process which works in two attempts....
neetu21 said:
Harendra ji, why not in 2 tries? Please explain :).
Click to expand...
 
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