A blindfolded man is asked to sit in the front of a carrom
board. The holes of the board are shut with lids in random
order, i.e. any number of all the four holes can be shut or
open.
Now the man is supposed to touch any two holes at a time and
can do the following.
? Open the closed hole.
? Close the open hole.
? Let the hole be as it is.
After he has done it, the carrom board is rotated and again
brought to some position. The man is again not aware of what
are the holes which are open or closed.
How many minimum number of turns does the blindfolded man
require to either open all the holes or close all the holes?
Note that whenever all the holes are either open or close,
there will be an alarm so that the blindfolded man will know
that he has won.
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Answer Posted / guest
The blindfolded man requires 5 turns.
1. Open two adjacent holes.
2. Open two diagonal holes. Now atleast 3 holes are open. If
4th hole is also open, then you are done. If not, the 4th
hole is close.
3. Check two diagonal holes.
o If one is close, open it and all the holes are open.
o If both are close, open any one hole. Now, two holes are
open and two are close. The diagonal holes are in the
opposite status i.e. in both the diagonals, one hole is open
and one is close.
4. Check any two adjacent holes.
o If both are open, close both of them. Now, all holes are
close.
o If both are close, open both of them. Now, all holes are
open.
o If one is open and one is close, invert them i.e. close
the open hole and open the close hole. Now, the diagonal
holes are in the same status i.e. two holes in one diagonal
are open and in other are close.
5. Check any two diagonal holes.
o If both are open, close both of them. Now, all holes are
close.
o If both are close, open both of them. Now, all holes are open.
| Is This Answer Correct ? | 8 Yes | 3 No |
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