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:

Fifteen Minutes.

After frying for five minutes, you can take out one steak and put the third one inside while turning the other one inside. After five more minutes, one steak is fried from both sides and you can take it out. Now put in the steak that we removed and turn the other one inside. After five more minutes, they both will also be fried from both sides.

Solution:

17 min

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 minutes. 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 minutes

Solution:

LOTTO

Explanation:
Just flip the two rows 1 and 2 vertically, you will find the word "Lotto".

Solution:

Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules

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:

Tilt the barrel until the fruit juice barely touches the lip of the barrel. If the bottom of the barrel is visible then it is less than half full.If the barrel bottom is still completely covered by the fruit juice, then it is more than half full.

Solution:

Playing chess

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

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:

9

Explanation:
2 < 3 < 5
To find out the required number of candies, take one in place of the least number (i.e. take one mango candy) and then add all the greater numbers (i.e. three strawberry and five pineapple candies) to it.

Required number of balls = 1 + 3 + 5 = 9.