Solution:

In total, five statements have been made in which nothing has been revealed by the possible culprit and they are A1, A2, A3, B3, and C1.

If you look closely at the statements A1 and C1, they are contradictory but that is not just it. It is clear that at most one of these statements can be true if both are not false.

Nothing much can be explained from the statements A2 and B3. But it is doubtful that A2 would be false and B3 would be true at the same time.

A3 is true as followed from the introduction.

Let us now focus on the remaining statements and see how many of them are true.

Statement B1
A is the thief: False
B is the thief: False
C is the thief: True
D is the thief: False
None of them is the thief: False

Statement B2
A is the thief: False
B is the thief: True
C is the thief: True
D is the thief: True
None of them is the thief: True

Statement C2
A is the thief: True
B is the thief: False
C is the thief: True
D is the thief: True
None of them is the thief: True

Statement C3
A is the thief: False
B is the thief: False
C is the thief: False
D is the thief: True
None of them is the thief: False

Statement D1
A is the thief: True
B is the thief: True
C is the thief: False
D is the thief: True
None of them is the thief: True

Statement D2
A is the thief: True
B is the thief: True
C is the thief: True
D is the thief: False
None of them is the thief: True

Statement D3
A is the thief: True
B is the thief: False
C is the thief: False
D is the thief: False
None of them is the thief: False

Now in total of all the statements
A is the thief: 4 True, 3 False
B is the thief: 3 True, 4 False
C is the thief: 4 True, 3 False
D is the thief: 4 True, 3 False
None of them is the thief: 4 True, 3 False

Now if we combine the results with the fact that A3 is correct, the results will change as
A is the thief: 5 True, 3 False
B is the thief: 4 True, 4 False
C is the thief: 5 True, 3 False
D is the thief: 5 True, 3 False
None of them is the thief: 5 True, 3 False

Now we know that only four statements can be true from all the twelve (the lie detector told us that). Thus statements A1, B2, B3 and C1 must be false in that case which means that Suspect B is the car thief.

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:

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:

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:

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:

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:

Put one black rock in one sack and put all the rest in the other one. In such manner you will get about 3/4 probability of surviving.

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:

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:

This problem can be solved with algebra as well as logical reasoning. We are going to tell you how it is logically possible.

Be it any quantity that was present in the beginning and any liquid as well. We know that we have taken a spoonful from one glass and put it into another. Then, we have taken a spoonful from the other glass and put it into the first glass. So, at the end of it, the quantity of liquid in either glass remains same as it was in the beginning.

Therefore, we can conclude the fact that any amount of coke that is missing in the glass with coke will be present in the lemonade glass. Also, the same quantity of lemonade will be missing from the lemonade glass and will be present in the coke glass.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.