Solution:

Were you thinking of using the fastest person who takes one minute to travel to and fro till everyone has reached the other end? Yes that can be done but it will take 10 + 1 + 7 + 1 + 2 = 21 minutes in total.

If we pay attention towards finding a way to merge the one taking 7 minutes and the one taking 10 minutes, we will have to acknowledge the fact that one of them will also have to return back which will take much more time than 21 minutes.

What if we use the person taking 2 minutes to escort the one taking 1 minutes across? Let's come to the conclusion:

1 and 2 cross the bridge
2 returns with the torch
7 and 10 cross the bridge
1 comes back
1 and 2 cross the bridge

In such a manner, the total time taken will be
2 + 2 + 10 + 1 + 2 = 17 minutes.

Solution:

The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.

Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.

Solution:

5 leopard and one sheep

since all animal are rational , once 1st leopard eats the sheep all the rest of leopard would know about this.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.

Solution:

A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)

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:

Put one red marble in one box and all other marbles in the 2nd box.

Probability :
50 + 24/49

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:

It can be down as shown below.

Solution:

It can be solve as shown below.