Solution:

4th Image

Images 1st 2nd & 3rd are images of the right hand while the 4th image is of the left hand

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 best thing for you will be to shoot your first shot in air.

If you try to shoot at Jack and hits him by chance, Leo always hits on the target and you are dead.

If you choose to shoot at Leo and hits him by chance, then there is a fifty percent chance that Jack will hit you.

If you shoot in air, the next turn is of Jack and he knows that Leo is a better shooter which means that he will shoot at him. If he misses, he is definitely dead as Leo will shoot on him coz he shot at him. The next turn will be yours and you stand a 1/3rd chance of hitting him. Also if, he is able to hit Leo. The next chance is yours again and you can try the 1/3rd probability of hitting over Jack.

This is the best situation that you can face. At every other possible way, you will face a worse situation for sure.

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:

It can be solve as shown below.

Solution:

Let us denote the number of boys as 'p' and the number of girls as 'q'.

The first statement says that each boy has equal numbers of brothers and sisters
(p - 1) = q
=> p - q = 1 ------- (a)

The second statement says that every girl has twice as many brothers as that of sisters
2 (q - 1) = p
=> 2q - p = 2 ------- (b)

By solving (a) and (b)
We get q as 3 and p as 4.

Therefore, there are 4 boys and 3 girls in the family.

Solution:

888 + 88 + 8 + 8 + 8 = 1000

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

192

Explanation:
I started with the number 192.
Adding the reversed 291 to 192 gave me 483.
Similarly,
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996

Solution:

The attached picture gives away the solution. If you can find another solution, do send us yours.

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.