Solution:

Did you consider the southern hemisphere? If you did you must be knowing that there is a ring near the South Pole with a circumference of one mile. Consider the situation that you are standing on one mile north of the ring at any point. Now if you walk on the southern direction and cover one mile, you will be standing on the ring. Traveling one mile east will bring you on the circumference of the ring. Walking one mile north from there, you will be standing on the exact point where you started. Now if you start counting, you will understand that while you walk 1 mile north in the end, you can reach an infinite number of points.

In such a case, the total number of possible points possible are 1 + infinite.

Now let us consider the ring that is half a mile in circumference near the South Pole. If you walk a mile along the ring, you would circle twice but will reach the point where you started from. In such a case if you start from the point that is located one mile north of a half mile ring, it will also help you reach the starting point after traveling as per asked.

Now with every possible integer N, there is a circle with radius R = 1 / 2 (2*pi*n); which is centered at the South Pole.

If you walk along these rings, you will be circling N times and again returning to the point where you started. You must note that the possible values for N are infinite. Also, you can have infinite ways of selecting a starting point which is located one mile north of the rings which means that there are (infinite * infinite) possible points.

Concluding with our statements, the possible number of points are equal to 1 + infinite * infinite which is equal to infinite.

Thus there are infinite points possible.

Solution:

Playing chess

Game needs two players so Brother-8 is playing chess with Brother-2

Solution:

Steps:
1) Light fire on first bar from both the ends and second bar from one end.

2) After 30 seconds(when first bar gets completely burned out) , burn the second bar from second end as well.

3) When second run completely gets burned , you know its 45 seconds.

Solution:

f the last elf is taking some time to answer the question then it shall mean that the elf's before him are all wearing distinct coloured hats. However sufficient time shall be given to the last elf to give the answer.
If he views the similar coloured hats in front of him , he shall quickly tell the answer

Solution:

Just One!

example
If u get Blue rock with label as "blue+red" => this is red as all labels are wrong.
In this way u can logically solve this puzzle.

Solution:

There can be two possibilities for the given situation.

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

Solution:

17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Solution:

The Lady sitting in the chair has the detector attached to the back of her blouse.

Solution:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

Solution:

Fifteen minutes can easily be measure using these two hour glasses.

Step 1: He will start both the hourglass.

Step 2: The moment the eleven minute hourglass is empty, he will invert it.

Step 3: When the thirteen minutes hourglass is empty, he will invert the eleven minute hourglass.
In step 3, we will have counted thirteen minutes. Since we inverted the eleven minute hourglass in step 2, it started from fresh and was inverted just for two minutes (13-11=2). In this manner when it is reversed when the thirteen minute hourglass is finished, it will have two minutes of sand left. This time when the sand finishes, he will have measured fifteen minutes. (13+2=15)

Solution:

It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.