THE OFFCICAL CAT2008: DI - lOGICAL REASOING

[srikar2097]

Originally Posted by PaglaMota

I don't remember when did CAT ask all these job puzzles in LR section. Do they really ask these type of Qs??

What "job puzzles" are you mentioning about? Could you please post their links here? Thanks...

[dare2]

Originally Posted by srikar2097

New LR Problem

This one is a very interesting. Have a crack at it

A king decides to give 100 of his prisoners a test. If they pass, they can go free. Otherwise, the king will execute all of them. The test goes as follows: the prisoners stand in a line, all facing forward. The king puts either a black or a white hat on each prisoner. The prisoners can only see the colors of the hats in front of them. Then, in any order they want, each one guesses the color of the hat on their head. Other than that, the prisoners can not speak. To pass, no more than 1 of them may guess incorrectly. If they can make their strategy before hand, the method they choose should save maximum prisoners. How many get saved? Explain the method.

99 of the prisoners can be saved for granted. the fate of the last prisoner will depend on his luck.

[prakharc]

hmmm....till now i can save 2 out of every 3
Last prisoner will shout white if both in front have white....will shout black if both in front have black..and will not say anything if one black and one white...So both in front of him will know their colour.....Now 4th person will do same for 5th and 6th.....So out of every 3 we will save atleast 2....fate of 3rd depends on luck

But yeah still thinking if we can save more.......

[srikar2097]

Originally Posted by prakharc

hmmm....till now i can save 2 out of every 3
Last prisoner will shout white if both in front have white....will shout black if both in front have black..and will not say anything if one black and one white...So both in front of him will know their colour.....Now 4th person will do same for 5th and 6th.....So out of every 3 we will save atleast 2....fate of 3rd depends on luck

But yeah still thinking if we can save more.......

Yupp, you can save more

[dare2]

Originally Posted by srikar2097

Yupp, you can save more

the last man will count the number of black caps in front of him. if odd number of black caps he can say black or if even then he'll say white.
the prisoner infront of him will then count the number of black caps in front of him and can very easily find out what colour cap he is wearing. in this way all can be saved except the prisoner standing last.

[prakharc]

Starting from last one:-

If both in front are white he'll shout white...If both in front black he'll shout black.....If different he wont say anything

Now prisoner in front of him will know his colour......From now on they will follow method.....If both in front have caps of same colour he'll shout his own colour.....else he'll remain silent signifying both in front have different colours.
So person in front of him will know his colour based on colour of the cap of the person standing in front of him....So this will go on till atlast only two remains....

Now if 3rd person shouted his colour that means 1st and 2nd have same colour..since 2nd knows the colour of the 1st he'll shout his colour and 1st one knows his is same as 2nd......If 3rd person doesn't shout his colour..that means 1 and 2 are different.....so after 2 shouts his own colour..1st knows his colour is different from 1

Thus everyone knows their colour correctly..Only last person might have done a mistake..but since 1 mistake is allowed..so everyone will be saved

[srikar2097]

Originally Posted by prakharc

Starting from last one:-

If both in front are white he'll shout white...If both in front black he'll shout black.....If different he wont say anything

Now prisoner in front of him will know his colour......From now on they will follow method.....If both in front have caps of same colour he'll shout his own colour.....else he'll remain silent signifying both in front have different colours.
So person in front of him will know his colour based on colour of the cap of the person standing in front of him....So this will go on till atlast only two remains....

Now if 3rd person shouted his colour that means 1st and 2nd have same colour..since 2nd knows the colour of the 1st he'll shout his colour and 1st one knows his is same as 2nd......If 3rd person doesn't shout his colour..that means 1 and 2 are different.....so after 2 shouts his own colour..1st knows his colour is different from 1

Thus everyone knows their colour correctly..Only last person might have done a mistake..but since 1 mistake is allowed..so everyone will be saved

Dude! You cannot keep quite You have to say the color of the hat. The method could be bettered to get the same result

[dhanu_25]

hmm..i think 99 can be saved..one has to sacrifice his life for his fellas..the first one can call out the color of the hat of the second one..and so on..there is a 50% chances that first one also gets saved..

[dare2]

Originally Posted by dhanu_25

hmm..i think 99 can be saved..one has to sacrifice his life for his fellas..the first one can call out the color of the hat of the second one..and so on..there is a 50% chances that first one also gets saved..

i think my solution has been overlooked. let me try to elaborate it this time.

let assume that there are 10 prisoners only. A,B,C,D,E,F,G,H,I and J in the order with A in front and J at last. let A,C,D,F,G and J have black caps on their head. and B,E,H and I have white caps.

now if they have decided that the last one will shout 'black' if there are even number of black caps before him else white, then

J - he will shout "WHITE" (J can see odd number of black caps before him excluding himself)

I - he'll count the number of black caps. since its odd he'll know that he is wearing white caps himself so he will shout white.

H - same as I

G - he'll count the number of black caps. since he can count 4 caps(even), he'll know that he is wearing black cap (because J, I and H have shouted white meaning odd number of black caps still remains including himself). so he'll shout 'black'

F - he'll count the number of black caps in front of him and the number of black caps that is already shouted (4 in all, even) so he should be wearing black cap. so he knows he should be wearing black cap because J shouted white meaning odd number of black caps. so he'll shout black.

and so on.. each will count the number of black caps in front of him and the number of black caps that is already shouted. and can very easily find the colour of the cap that he is wearing..

[Purvika]

Originally Posted by IIM maniac

@ Implex , exactly that what i did, so its confirmed that German keeps
the FISH.
@ iwillwin, there is indeed a point in your saying, but i think its correrct if
we assume the left pet is FISH.
PUYS more inputs??

hiiiiiiiii dis is purvi can give some free advice if possible plzzzzzzzz