Solution:

Password

Solution:

You must put labels on the tablets as A and B before using. In that case, if you pour tablets together, you will get 3A, 2A 1B, 1A 2B or 3B. If they are from the same bottles you can take one from another bottle and save the remaining two for another day. If you get two from same and one from other, you can draw one from another bottle and you will have two pairs of which you can eat one and save the other.

Solution:

3/7

Explanation:
If a 31day month starts on a Friday, Saturday or Sunday it will have five Sunday, if not it will have 4 Sunday.

Solution:

Flyball should pay Lavy $11 instead of accepting his counter offer as in that case the club will end up paying almost double the amount.

Solution:

The competition was held on 11 November 2013. The logic is simple. In America, the dates are mentioned as Month/Day/Year and in Britain, dates are written as Day/Month/Year. But on 11 November 2013, the date and month will be the same. And thus both will write in the same manner i.e. 11/11/2013. If it would have been any other date, the evaluators would have been able to identify it.

Solution:

Align the toothpicks to form TEN. Use one toothpick for each of the lines.

Solution:

Costa

Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f

=> dqvxf = costa

Solution:

Since each one of them gave one answer wrong, this means that each one of them gave two answers right.

Let us assume that Student A gave a wrong answer to the first question. This will mean that Student B also gave a wrong answer for the first. This will conclude that the rest of the two answers given by them are correct. However, the answers are different and thus it is not possible.

Thus both Student A and Student B must be right with the first question and the answer to the first is two.

If you keep applying the same logic, you will come to a conclusion that following are the correct answers:
Q1. Two
Q2. Three
Q3. Two

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:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

888 + 88 + 8 + 8 + 8 = 1000