Solution:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver

Solution:

It was only a trick

I borrowed my parents (Pound 50 + Pound 50 = Pound 100).
I bought the book for Pound 97 and i had Pound 3 change.
So, the money that i borrowed and bought the book are Pound 100 (borrowing money)
and the change Pound 3 also the borrowing money the book is Pound 97
Total: Pound 97 + Pound 3(the change) = Pound 100
The money that i used to buy the book all i borrowed.
Pound 97 is the borrowing money, change: Pound 3 also the borrowing money. Pound 97 + Pound 3 = Pound 100

Solution:

40 balls

Solution:

NONE, holes are empty.

Solution:

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

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:

XIX is the answer which refers to nineteen in roman numeric system. On taking out one i.e. I, we will get XX that is equal to twenty.

Solution:

Here is a way to do it. We dont know if it can be done using some other arrangement or not. If you know, do send us.

4 | 3 | 8
9 | 5 | 1
2 | 7 | 6

Solution:

There can be two possibilities for the given situation.

One is if I run on the perimeter. Then, the lion will eventually catch me as the lion can follow my radius and then his trajectory will be half of the circle and not a spiral. Therefore, the lion will subsequently catch me in a finite amount of time.

Secondly if I don’t run on the perimeter, then clearly, I have an infinite amount of time before the lion catches me.

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:

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.