Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.
At a restaurant downtown, Mr. Red, Mr. Blue, and Mr. White meet for lunch. Under their coats they are wearing either a red, blue, or white shirt.Mr. Blue says, 'Hey, did you notice we are all wearing different colored shirts from our names?' The man wearing the white shirt says, 'Wow, Mr. Blue, that's right!'
Can you tell who is wearing what color shirt?
John is out with his class of 25 boys to a local park. Each guy has a remote controlled car with them. The park has a racetrack that allows 5 cars to be raced at once. Their teacher, Mr. Ted, declares that the top three fastest cars get ice cream.
How many races are required to determine the 3 fastest cars?
Your friends are coming over to your birthday party. Your mum has bought this delicious giant doughnut for you guys. Unfortunately she had to go somewhere. Now you have to serve all of them including you.
There are a total of nine kids including you. None of you would mind a smaller piece. Can you cut this doughnut into nine pieces in just three straight cuts?
At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple
Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.
If we tie a sheep to one peg, a circled grass is been eaten by the sheep. If we tie the sheep to two pegs with a circle on its neck, then an eclipse is eaten out of the grass by the sheep. If we want an eclipse then we put two pegs then put a rope in between then and the other end of the rope is tied up on the sheep's neck.
Question: how should we tie the peg and the sheep so that a square is eaten out from the garden's grass? We only have one sheep's rope and the peg and the rings.
This is a famous probability puzzle in which you have to choose the correct answer at random from the four options below.
Can you tell us whats the probability of choosing correct answer in this random manner.
You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.
The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.
Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.
So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.
There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3
(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.
That was the easy part.
What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.
Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.
Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.
For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.
In the given picture, you can find a few numbers. Now you have to fill each square of the grid in a manner that every row and every column contains the digits 1 to 6. Another thing to keep in mind is that the connected squares must have the same number in them.