Solution:

28 =>1+2+4+7+14

cannot be 1 (as number cannot be included)

Solution:

There are six unique answers

Ans1:
7 x 4 = 28
36 / 4 = 9
1 + 4 = 5

Ans2:
9 x 4 = 36
28 / 4 = 7
1 + 4 = 5

Ans3:
3 x 8 = 24
56 / 8 = 7
1 + 8 = 9

Ans4:
7 x 8 = 56
24 / 8 = 3
1 + 8 = 9

Ans5:
4 x 8 = 32
56 / 8 = 7
1 + 8 = 9

Ans6:
7 x 8 = 56
32 / 8 = 4
1 + 8 = 9

Solution:

The Password.

Explanation:
Clues 2 and 5 are correct and rest are lies.

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:

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

Probability :
50 + 24/49

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

Solution:

I filled a glass full of water and held it over my head for 20 minutes.

Solution:

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

Solution:

There are 3 fruits (1 grapes, 1 apples, and 1 oranges).

Solution:

The check was for dollars 31.63.

The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.

Let x be the dollars of the check, and y be the cent.

The check was for 100x + y cent
He was given 100y + x cent

Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5

or 199x ≡ -5 (mod 98)
or 98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
this quickly leads to x = 31

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.