Solution:

10 Bulbs

For this tricky puzzle, you must check how many light bulbs in the row are having an odd number of factors. The first one surely has odd number of factors, the second has even, four has odd. Thus the bulb four and one will remain lit. The bulbs that are going to remain lit are perfect squares as they have an odd number of factors - 1, 4, 9, 16.

Since there are 100 passes, you can go up to 10 times 10 i.e. the square of 10. There are 10 perfect squares available to you - one, two, three, four, five, six, seven, eight, nine and ten. They corresponds to the bulb number 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. All of them will remain lit and thus total ten bulbs will remain lit after 100 passes.

Solution:

The question can practically disarm you. But we promise by the time you read the solution, you will find that it was way easy than you were thinking.

Let us name the boxes with a number - Box1, Box 2..., Box10. Now you must be familiar with the weights of the balls precisely. By saying that we are not implying that you will have to take one ball from every box and judge the weight of every ball.

What you have to do is pick one ball from Box 1, 2 balls from Box 2, 3 balls from Box 3, ...., 10 balls from Box 10. Thus after taking balls from every box in such a manner, you will finally have 55 balls in total. There comes the time to use the weighing machine.

If all the balls were weighing adequate, the combined weight should be 55 x 10 grams = 550 grams. But since one of the box has defected balls, the weight will be less than that. Here is the tricky part and explains why we took different number of balls from each box.

If the total weight is less than 1 gram, then the defective box is the Box 1 since we took 1 ball from that box. If the Box 2 is defective, the total weight will be less than 2 grams. In similar fashion you can identify the defective box by analyzing the total weight that is calculated by the weighing machine.

Solution:

Seven

Dividing the horses in a group of five, we will have five different groups of horses. A race will be conducted separately for each group which will help us find the fastest three horses in every group. We have used five different races till now.

Now we can have a race among the winners from every race which will help us determine the fastest of them all. We have used the sixth race.

The second fastest horse can be the one who came second in the sixth race or the second fastest horse in the finals winners' first race.

If we look at the first case, the third fastest horse can be the third fastest horse in the sixth race or the second fastest horse in either the fastest or second fastest horse first race.
If we look at the other case, the third fastest horse may be the second fastest in the sixth race or the third fastest in the final winners' first race.

If you calculate our assumptions, we have five more horses to race (seventh race) and determine the second fastest and the third fastest horse.

Solution:

Just One!

example
If u get Blue rock with label as "blue+red" => this is red as all labels are wrong.
In this way u can logically solve this puzzle.

Solution:

He should shoot at the ground.

If Kangwa shoots the ground, it is Rafael's turn. Rafael would rather shoot at Ferdinand than Kangwa, because he is better.
If Rafael kills Ferdinand, it is just Kangwa and Rafael left, giving Kangwa a fair chance of winning.
If Rafael does not kill Ferdinand, it is Ferdinand's turn. He would rather shoot at Rafael and will definitely kill him. Even though it is now Kangwa against Ferdinand, Kangwa has a better chance of winning than before.