Solution:

B, D and E are absolutely identical to the green figure.

Solution:

2 people

Say 2 helpers are Helper1 & Helper2

Helper1 food Helper2 food McDuck's Food
End of Day 1 : 1 4 4
Now 'A' will head back

End of Day 2 : N/A 2 4
Now 'B' will head back

Now Mr. McDuck have 4 day food :-)

Solution:

2 degree below zero

Solution:

Hockey Stick $49.50 & ball $0.50

Solution:

He should draw any chit of paper and before unfolding it, he must eat and swallow the chit. In this manner, to determine the judgment of the god, they will have to look at the other chit which will have 'Guilty' written over it. Thus everyone will believe that the chit which he ate had innocent written over it and thus they will believe that he is innocent.

Solution:

13 + 11 + 6 = 30

Explanation
The trick is that you just need to rotate '9' to form '6'.

Solution:

If you ask the guard who always tells the truth, he would tell you the other guard would point you to the door of death. If you ask the guard who always lies, he would tell you the opposite door of the truth-telling guard and point you to the door of death. In either case, both guards will point to the door of death so you should choose the other one.

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:

The teacher will use a simple trick to form a perfect impression. He will ask all his students to raise their hands on each of the question that is asked. But the only catch will be that those who knows the answer correctly will raise their right hand and others will raise their left hand.

In this way, the principal will see all the hands being raised for each question even though all won't be knowing the correct answer. The teacher will ask only those who know the answer and they will always be correct. So the principal will be impressed to full extent.

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:

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