Solution:

Andrew had been teaching 21 years.
Beyonce had been teaching 15 years.
Curt had been teaching 7 years.

Now let us consider the data given to us in the question
Andrew and Beyonce had been teaching for a total of 36 years.
Curt and Beyonce had been teaching for a total of 22 years.
Curt and Andrew had been teaching for a total of 28 years.
Let Andrew = A, Beyonce = B and Curt = C, then:
A + B = 36 [1]
C + B = 22 [2]
C + A = 28 [3]

If we use [3] - [2] we have:
A - B = 6 [4]

If we use [1] + [4] we have:
2A = 42
A = 21

By [1] B = 15
By [3] C = 7

Solution:

12 steps:

Steps 11 Liter jug 6 Liter Jug
1. 11 -
2. 5 6
3. 5 0
4. 0 5
5. 11 5
6. 10 6
7. 10 0
8. 4 6
9. 4 0
10. 0 4
11. 11 4
12. 9 6 ==> i found 9 liter in 11liter jug(as required)

Solution:

Sansa has the siamese (conjoined) twin sister.

Solution:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

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:

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:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

0%

Explanation:
1) why cant be 1/4 : If the answer is 1/4, then as we know two out of four answer choices is '1/4', the answer has be 1/2.
This is a contradiction, so the answer cannot be 1/4.

2) why cant be 1/2 : If the answer is 1/2 then because answer:"1/2" is 1 out of 4 answer choices, the answer must be 1/4. This is also a contradiction. So the answer cannot be 1/2.

3) why cant be 1 : If the answer is 1 then because answer:"1" is 1 out of 4 answer choices, the answer must be 1/4. Again the same contradiction and therefore answer cannot be 1.

Solution:

It can be down as shown below.

Solution:

99 matches.

The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 99 matches were played.

Solution:

The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.

Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.

In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.

Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.