Solution:

when you are a mouse.

Solution:

A typical way will be to start thinking about summing up the distance that the bee will travel but that will be quite a tedious task. How about we offer you a much easy solution?

The tunnel is 200 miles long and the trains are running at as peed of 100 mph which means that they will collide exactly at the center of the tunnel and seeking their speed, they will collide after an hour.

Now consider the bee which is flying at a speed of 1000 mph and will keep flying till the train collides. As calculated, it will keep flying for an hour which means the distance that it will cover is 1000 miles.

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

The stranger said that he thought the room was his. If so, he must have had the keys to the room and thus he must not have knocked.

Solution:

David to play second

Explanation:
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.

Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.

If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.

Therefore, you should suggest David to play second.

Solution:

Numbers of the first cube:
1, 2, 4, 0, 5 and 6.

Numbers on the second cube:
3, 1, 2, 7, 8 and 0.

Now you can pick any date and you will be able to show it using these cubes. Remember that you can always use 6 as 9 by inverting it.

This problem has another solution and we leave it for you to deduce.

Solution:

The clerk said that every day the password is CHANGED. Changed is the password.

Solution:

Kid1, he was left with one rock.

Solution:

TIF56

Explanation:
The password is formed as:
T (5th letter of Shruti)
I (5th letter of Arshika)
F(1st letter of Friday)
5(5th day of week)
6(Length of Friday)

Similarly, we will pick the 1st letter for Monday and 3rd letter for Wednesday.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

Old password : altered

New Password : distinct