Solution:

The cheerleaders.

Solution:

A Spider

Solution:

Selfie
Explanation:
: Selfie = is six letter word
: Self = The first four letter is me.
: e = The second and last letter are the same.
: fee = The fourth second and last letter are payment.

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:

Are you Thorgan

Explanation:
Below will be the reply of three brothers:
Eden: No (speaks truth)
Thorgan: No (speaks lie)
Kylian: Yes (speaks lie)

Therefore if its Kylian then only you will get a reply as "Yes" and then you ask for your money.

Solution:

Remove the same 2 matchsticks from each figure and move figure towards left 45 degrees, It forms the letters V W X Y.

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:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

The teacher will use a simple trick to form a perfect impression. He will ask all his students to raise their hands on each of the question that is asked. But the only catch will be that those who knows the answer correctly will raise their right hand and others will raise their left hand.

In this way, the principal will see all the hands being raised for each question even though all won't be knowing the correct answer. The teacher will ask only those who know the answer and they will always be correct. So the principal will be impressed to full extent.

Solution:

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

Probability :
50 + 24/49

Solution:

The young lady was presented with a bill which belonged to the elderly woman. She had asked the waiter to collect the bill from her daughter.