Solution:

Kid1, he was left with one rock.

Solution:

13 x 3 + 1 = 40

The number of decks is absolutely not relevant here which means whether we mix 5 or 500 cards still results would be same.
Any card drawn will be a Ace,2,3,4,5,6,7,8,9,10,Jack,Queen or King, so there are 13 possibilities each time a card is drawn.
If you are lucky just 5 cards of the same kind can be obtained in 4 steps

The unluckiest(worst) way is our solution as we need to guarantee a four of a kind.
Draw 3 of each kind =>now we have 39 cards. Next card will guarantee 4 of a kind.

Solution:

If two widows, each having a son married the son of the other widow, and then by the consummation of marriage, both the couples had a daughter, all the aforementioned relationships will turn to be true.

Solution:

I am Spice

Explanation:
Spice is hot if you remove first two alphabets then it will become "ice", which is really cold.

Solution:

4 6 5 3 (from magician perspective)
3 5 6 4 (viewer perspective)

Detail:

From user perspective:
3 5 6 4 => 5 6 4 3 => 6 5 4 3

For magician perspective
3 4 5 6 (ascending order as)

Solution:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

Solution:

The bridge won't break.

When the car has travelled half of the bridge, that is 15km, the car must have used more than 300 grams of fuel. Therefore, the impact of bird's added weight will be zero literally.

Solution:

2 cakes

At each toll bridge you are required to give half of your cakes, and you receive one back. Which leaves you with 2 cakes after every bridge.

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.

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:

Michelangelo juggles the diamonds across while crossing the bridge.