You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls.
You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls.
What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other.
If the scale is balanced, that means the odd ball out is in the other 4 balls.
Let's call these 4 balls O1, O2, O3, and O4.
Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale.
If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier.
If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter.
If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter.
If the scale isn't balanced, then the odd ball out is among these 8 balls.
Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier").
Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter").
Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls.
So now weigh [H1, H2, L1] against [H3, L2, Normal].
-If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter.
-So measure [H1, L2] against 2 of the "Normal" balls.
-If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier.
-If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter.
-If the scale is balanced, then H2 is the odd ball out, and is heavier.
-If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier.
-So measure L1 and H3 against two "normal" balls.
-If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter.
-Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier.
If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter.
So weight [H4, L3] against two of the "Normal" balls.
If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier.
If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter.
If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
Tarun Asthnaiya go to his office by local train. However nearby train station is quite far from his place and he used to drive his bike to train station daily with an average speed of 60km/hr. One day at halfway point he relized that due to heavy traffic he got late having average speed of just 30km/hr. How fast he must drive for the rest of the way to catch my local train?
The train is just about to leave the station and there is no way Tarun will be able to catch it this time.
A man needs to send important documents to his friend across the country. He buys a suitcase to put the documents in, but he has a problem: the mail system in his country is very corrupt, and he knows that if he doesn't lock the suitcase, it will be opened by the post office and his documents will be stolen before they reach his friend.
There are lock stores across the country that sell locks with keys. The only problem is that if he locks the suitcase, he has no way to send the key to his friend so that the friend will be able to open the lock: if he doesn't send the key, then the friend can't open the lock, and if he puts the key in the suitcase, then the friend won't be able to get to the key.
The suitcase is designed so that any number of locks can be put on it, but the man figures that putting more than one lock on the suitcase will only compound the problem.
After a few days, however, he figures out how to safely send the documents. He calls his friend who he's sending the documents to and explains the plan.
What is the man's plan?
The plan is this:
1. The man will put a lock on the suitcase, keep the key, and send the suitcase to his friend.
2. The friend will then put his own lock on the suitcase as well, keep the key to that lock, and send the suitcase back to the man.
3. The man will use his key to remove his lock from the suitcase, and send it back to the friend.
4. The friend will remove his own lock from the suitcase and get to the documents.
Search: Man-in-the-middle attack
A man was in a small town for the day, and needed a haircut. He noticed that there were only two barbers in town, and decided to apply a bit of logical deduction to choosing the best one. Looking at their shops, he saw that the first one was very neat and the barber was clean shaven with a nice haircut. The other shop was a mess, and the barber there needed a shave and had a bad cut besides. Why did the man choose to go to the barber with the messy shop?
Since even barbers rarely try to cut their own hair, and there are only two barbers in town, they must cut each other's hair. The one with the neat hair must have it cut by the one with the bad haircut, who must then be better one, considering his own haircut.
Why are manhole covers round? Do manhole covers really need to be circular?
Manhole covers are round so that they won't fall through the hole into the sewer below them. No matter how you turn the cover, you won't be able to push the cover through the hole.
However, if you were to have square manhole covers, you would be able to rotate the cover such that one of the edges of the square cover is lined up with the diagonal line of the square hole, which would allow the cover to fall through, causing countless problems that the general public would rather avoid.
Many years ago a wealthy old man was near death. He wished to leave his fortune to one of his three children. The old man wanted to know that his fortune would be in wise hands. He stipulated that his estate would be left to the child who would sing him half as many songs as days that he had left to live.The eldest son said he couldn't comply because he didn't know how many days his father had left to live and besides he was too busy. The youngest son said the same thing. The man ended up leaving his money to his third child a daughter. What did his daughter do?
Every other day, the daughter sang her father a song.
A murdered is condemned to death.
He has to choose between three rooms.
The first is full of raging fires, the second is full of assassins with loaded guns, and the third is full of lions that haven't eaten in 3 years.
Which room is safest for him?
The third room. Lions that haven't eaten in three years are dead.
You can easily "tile" an 8x8 chessboard with 32 2x1 tiles, meaning that you can place these 32 tiles on the board and cover every square.
But if you take away two opposite corners from the chessboard, it becomes impossible to tile this new 62-square board.
Can you explain why tiling this board isn't possible?
Color in the chessboard, alternating with red and blue tiles. Then color all of your tiles half red and half blue. Whenever you place a tile down, you can always make it so that the red part of the tile is on a red square and the blue part of the tile is on the blue square.
Since you'll need to place 31 tiles on the board (to cover the 62 squares), you would have to be able to cover 31 red squares and 31 blue squares. But when you took away the two corners, you can see that you are taking away two red spaces, leaving 30 red squares and 32 blue squares. There is no way to cover 30 red squares and 32 blue squares with the 31 tiles, since these tiles can only cover 31 red squares and 31 blue squares, and thus, tiling this board is not possible.
