Logic Riddles

The probability will be 1/36.

For the first dice, there can be six possibilities. Similarly, for the second dice as well, there can be six possibilities.
Thus the total possibilities is 6 * 6 = 36.
The outcome we need is that the first comes up with a 2 and the second comes up with a 5. That is possible only in one possibility.
Therefore the required probability is 1/36.

There are three flowers in the pond and four bees are hovering over it.

This puzzle is known as the Turan Brickyard Problem and is a classic one. There is no way that they can be connected without crossing the lines in 2 dimension. However, you can do it if you place the scene in a 3d workplace. See to the figure to understand better.

If we are considering an ideal case and we have to calculate the maximum score, we will assume that the batsman hits six on every ball. But then, at the end of every over, the strike will change. But we have to make sure, he gets to play maximum number of balls to score maximum runs. Thus, the batsman will hit six in first five balls and will choose to take 3 runs in the last ball so that he retains the strike in the following over.

Total runs per over in that case = 6 * 5 + 3 = 33.

But in the last over (50th over), he don’t have to worry about keeping the strike and thus, he will hit six even on the last ball. Therefore, the maximum runs that a batsman can score:

33 * 49 + 6 * 6 = 1617 + 36 = 1653

The back seat of the last coach will be the best. The reason is because, the train will be accelerating after the stop and thus, it will be much faster when the back of the train enters the tunnel than when the front of the train enters. Thus you will have to spend less time in the tunnel.

It is possible if you are using Roman Numerals.
9 = IX
10 = X
11 = XI
Now it is all possible.

146900

In such cases, we calculate from the last thing. Thus in this case, let us start counting from the 9th temple. Since, he must have offered 100 coins while climbing down stairs, it means that he must have offered Rs. 100 to god on the ninth temple and offered Rs. 100 while climbing up steps. Thus it is clear that he had Rs. 300 in his pocket before climbing the steps of the 9th temple.

Similarly, we will calculate for all the temples backwards.
Before eight temple: (300+100)*2 + 100 = 900
Before seventh temple: (900+100)*2 + 100 = 2100
Before Sixth temple: (2100+100)*2 + 100 = 4300
Before fifth temple: (4300+100)*2 + 100 = 8900
Before fourth temple: (8900+100)*2 + 100 = 18100
Before third temple: (18100+100)*2 + 100 = 36,500
Before second temple: (36500+100)*2 + 100 = 73300
Before first temple: (73300+100)*2 + 100 = 146900

Therefore he had Rs. 146900 initially.

Suppose you are player 1 and your friend is player 2.
You pick up 9, then your friend will choose 8.
Now you won’t be able to pick up 7 as then your sum will be more than 15 which will make you lose. But then your friend will pick 8 and will win.

Thus you will start by picking 1. If your friend picks up 2, then you will pick 3 and your friend will pick 4. Now this will force you to pick 9. The score now becomes 6 to 13 and you have no chance of winning. Thus you pick up 9 after your friend picks up 2. Then the player 2 will pick 8. The score will now be 10 to 10. Thus you pick the number 3 as picking 7 will throw him over 15. Your friend will pick 4. Now you will have no move, thus your friend will win. So it will be wise to play second.

Simple answer :=)

Whenever you are drawing the balls, the number of red balls can decrease by either 2 or not decrease at all. However, in the case of blue balls, at each drawing, they can either go down by 1 or increase by 1.

Therefore, if the outset is assumed to begin with at least one ball in the urn to begin with and the number of red balls are 0 or even in number, all the red balls will finish and a blue ball will remain at the end. In the other case, one red ball will remain.