Solution:

The five cards must contain at least one card of the same suit.
Let's assume Rhodes placed one of the two cards of the same suit at the end and the second card was given to Daniel.
In that case, Daniel will know the suit of the card.

Now the real question is how Rhodes made sure that Daniel also knows the number of the card.

Daniel got four cards and one of them will determine the suit, so we are left with 3 cards.
3 cards can be arranged in 3! ways i.e. 6 ways but our card numbers can vary from 1 to 13.
Since we have two cards of the same suit we will make sure that we don't pick the King (13). This leaves us with 12 numbers.

We can also distinguish card from upside down position, therefore, we now have 6 * 2 ways to arrange these card.
"TADA.... We cracked the magic trick".

Let us assume that the smallest card is C1U (upwards position) and C1D (downwards position). Similarly, we have C2U, C2D, C3U and C3D

According to that, the 12 possible arrangements are:
C1U C2U C3U => NUMBER 1
C1U C3U C2U => NUMBER 2
C2U C1U C3U => NUMBER 3
C2U C3U C1U => NUMBER 4
C3U C1U C2U => NUMBER 5
C3U C2U C1U => NUMBER 6
C1D C2D C3D => NDMBER 7
C1D C3D C2D => NDMBER 8
C2D C1D C3D => NDMBER 9
C2D C3D C1D => NDMBER 10
C3D C1D C2D => NDMBER 11
C3D C2D C1D => NDMBER 12

Please note that king (number 13) is not possible as discussed already.

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:

Baseball umpire and the Catcher (Eden Hazard was playing baseball.)

Solution:

The woman told him to take out one nut each from the rest of the wheels and use them.

Solution:

The wise man told them to switch the horses and then race. In this manner, both of them will be willing to race as fast as they can to prove that the other horse (his own horse) is slow.

Solution:

Six.

There are only five colors, so six socks guarantee that two will be the same color.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

4

No Of circles in the number.

Solution:

Four

three placed flat on the table in a triangle(touching each other) and put the fourth one on top of them in the middle.

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:

It can surely be true because it is not mentioned that it was raining.