"Welcome back to the show. Before the break, Mr Ixolite here made it to our grand finale! How do you feel Mr.Ix?"
"Nervous."
"Okay, now to win the star prize of one million pounds all you have to do is answer the following question in 90 seconds."
"Okay, I'm ready."
"Right. In 90 seconds name 100 words that do NOT contain the letter 'A'. Start the clock!"
Can you help?
One, two, three, four, five...one hundred! I just counted from 1 to 100 in ninety seconds (it is possible).
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.
A boy was at a carnival and went to a booth where a man said to the boy, "If I write your exact weight on this piece of paper then you have to give me $50, but if I cannot, I will pay you $50."
The boy looked around and saw no scale so he agrees, thinking no matter what the carny writes he'll just say he weighs more or less. In the end the boy ended up paying the man $50. How did the man win the bet?
The man did exactly as he said he would and wrote "your exact weight" on the paper.
I run over fields and woods all day. Under the bed at night I sit, never alone. My tongue hangs out, up and to the rear, waiting to be filled in the morning. What am I?
It was a very large truck. The truck need to cross a 20 mile long bridge. Unfortunately, the bridge can only hold the weight of 12000 lbs. Even a single pound extra, the bridge would collapse. However the weight of the truck is exactly 12000 lbs. The driver carefully drove and crossed almost 85 percent distance of the bridge. He stopped to get a small break. Suddenly, a bird landed on the truck. Did the bridge collapse? Justify your answers with explanation!
No. The bridge doesn't collapse. The truck almost crossed 85 percent of total distance. Equivalent diesel would have been lost. So the extra weight of the bridge doesn't add any extra load to the bridge.
