Solution:

405

For an explanation, watch the video.

Solution:

Ms. Britney did it.

We will take up every person and analyze the statements over all of them one by one.

Case 1: Ms. Britney did it, then
Statement 1 => true
Statement 2 => false (as Rihanna did not do it)
Statement 3 => false (as Ms. Britney is lying)
Statement 4 => false (as Ms. Gwyneth is lying)
Our first guess itself was right but let us confirm our analysis by analyzing the statements over others.

Case 2: Ms. Rihanna did it, then
Statement 1 => false (as Ms. Britney did not do it)
Statement 2 => true
Statement 3 => true (as Ms. Britney is not lying)
As per the conditions, three statements had to be wrong but here, we have two statements right which clearly contradict the condition.

Case 3: Ms. Gwyneth did it, then
Statement 1 => false (as Ms. Britney did not do it)
Statement 2 => false (as Ms. Rihanna did not do it)
Statement 3 => false (as Ms. Britney is lying)
Statement 4 => false (as Ms. Gwyneth is lying)
She did not do it as all the statements are false.

Case 4: Ms. Yoon did it, then
Statement 1 => false (as Ms. Britney did not do it)
Statement 2 => false (as Ms. Rihanna did not do it)
Statement 3 => false (as Ms. Britney is lying)
Statement 4 => false (as Ms. Gwyneth is lying)
All four statements are false in her case as well thus she did not do it.

Therefore we have proved that Ms. Britney did it.

Solution:

There is no mention if you can use sign or not and thus the only possible way to get this is as follows:

(15 - 9) + (13 - 7) + (7 - 1) + (9 - 1) + (13 - 9)

If you solve inside the brackets, you will get the following equation
6 + 6 + 6 + 8 + 4
Adding all these numbers, you will get 30 which is the required sum.

Solution:

Firstly: Policeman should take the thug across the river thames and escort the boat back with him. (Thug across, everyone else on this side)
Then, Policeman will take one cat along, leave it there, and on the way back, bring back the thug (since thug cannot be left with anyone without policeman's supervision).
Then Mrs. Mother will take the second cat on the other side. Drop it there and bring the boat back. For the 4th trip, Mrs. Mother And Mr. Father Will travel to the other side, and Mrs. Mother will stay there, and MR. Father will bring the boat back. (Now Mrs. Mother and 2 cats are on the other side.)
5th trip, the policeman takes the thug across. Both of them stay there and Mrs. Mother will come back. Then for the next trip, Mr. Father and Mrs. Mother will go on the boat, and Mrs. Mother will stay there, and Mr. Father will return back.
Then for the 7th trip, Mr. Father will travel with one dog. On the return, the policeman and the thug will return. Then in the next trip, the Policeman will take the second dog. (Now Mrs. Mother and her two cats; Mr. Father and his two dogs and the Policeman is here).
Now finally the policeman takes the boat on the other side and brings back the thug.
Tough one, right ?

Solution:

Zouma is 3 and has two footballs.
Ake is 1 and has one football.
Kalas is 4 and has three footballs.
Aina is 2 and has four footballs.

Solution:

The Password.

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

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:

See the colored sticks in the picture. We have moved those sticks in the position given in the second image corresponding with the color.

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

A Question

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.