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:

We agree that the series is quite difficult to crack and much tricky as well.

1 can be written as one (consist of 3 letters)
2 can be written as two (consist of 3 letters)
3 can be written as five (consist of 5 letters)
4 can be written as four (consist of 4 letters)
5 can be written as five (consist of 5 letters)
6 can be written as six (consist of 3 letters)

Thus the next term in the series is 3.

Solution:

What we will do first is we will subtract the driving time of twenty minutes from the 2 hours 30 minutes difference and we are left with 2 hours 10 minutes.

Now Tia left in morning, thus for the mirror effect, the hands must be pointing up and down i.e. at 6 o'clock. (It could have been at 12 o'clock as well as the hands are straight up but we know she left in the morning).

Now we have to divide 2 hours and 10 minutes by 2 in order to find a center point that will be 65 minutes in this case for compensating for the mirror.

Now if we add 65 minutes to 6 o'clock, we get the time when she left the home which will be 7:05 and the time which she saw in the mirror was 4:55.

Now if we add the 20 minutes we had subtracted previously again to the time she left her home, we will get the time she reached at the party which will be 7:25.

Thus the time when she reached the party was 7:25.

Solution:

Put five chocolates in the first bag, two in the second and three in the third.

Now insert the bag containing three chocolates in the bag containing 2 chocolates.

Now, you have one bag with five chocolates, another bag with five chocolates and the third bag with three chocolates which are each odd in number.

Solution:

Let us assume that the z axis denotes the time and x and y axis denote the 2D planet. The four trajectories in that case are straight lines. Since no collisions involved more than two air planes, the four lines must be lying in the same plane. Now you must be thinking that the other two planes will collide as well.

But consider the fact that they might have collided in the negative time. Thus you can't be sure with the provided facts if the two remaining planes will collide or not.

There can be both possibilities. The first assumption depicts that the four air planes fly in a manner that there is no collision between the ships. However in the second diagram, the four airplanes fly in a manner that there is collision between them.