Solution:

developers=2, testers=5 and content-writters=13

Explanation:
Let the number of developers, testers, and content-writers are denoted by d, t, and c respectively.

From the given information we can write the following 2 equations;
d + t + c = 20 . . . . . . (1)
3d + 1.5t + 0.5c = 20 . . . . . (2)

Multiplying equation (2) by 2 we get;
6d + 3t + c = 40 . . . . . (2)

equation3 - equation 1
5d + 2t = 20

The unique solution with whole numbers is d = 2 & t = 5

Therefore from equation(1) we can find;
c = 13

developers=2, testers=5 and content-writters=13.

Solution:

At 01:05

Solution:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

Both will rotate at Same speed.

Explanation:
Gears of the same size will always rotate at constant speeds despite any smaller or larger gears in between them.

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.

Solution:

David to play second

Explanation:
Let's suppose that David plays first and he picks 9. Then Albert will definitely pick 8. Now, David will have to pick 7 or Albert will pick 7 in his turn. But if David picks up 7, then he will score 16 that is beyond 15 and will lose. So one thing is for sure, no one will be willing to start with the highest digits.

Suppose David plays first and picks up 1, Albert will pick 2. Then David will pick 3 and Albert will pick 4. Now David will be forced to pick 9. The score is 6 to 13 and thus David will have no chance of winning.

If David Picks 9 after Albert has picked up 2, then Albert will pick 8 and the score will become 10 to 10. Thus David will pick 3 as picking 7 will send him past 15. Now Albert will pick 4 and David has nothing to pick for winning. Thus Albert wins.

Therefore, you should suggest David to play second.

Solution:

The 100th statement for sure is incorrect because it says that at least 100 of the statements are incorrect.

Suppose if that is correct, then 100 statements cannot be true.
This suggests that the 100th statement is incorrect and that the first statement is true.

Similarly 99 statements cannot be true because if they were true, then two statements would become correct i.e. the 1st and the 99th.
But the 99th statement says that at least 99 are incorrect.
This suggests that the 99th statement is incorrect and that 2ndone is true.

If we keep analyzing is the same way till the end, we will find out that only the first fifty statements are true and all the remaining ones are incorrect.

Solution:

Put one red marble in one box and all other marbles in the 2nd box.

Probability :
50 + 24/49

Solution:

When you were inside the shed, the murderer replaced the lock with his own one that was identical to your original lock. When you locked the store room, you did not require a key and there was nothing abnormal for you. When you left, the murderer opened the lock with his key, planted the dead body inside and replaced the lock again putting the original lock in the place. He then closed it without any problem.

Solution:

I am Spice

Explanation:
Spice is hot if you remove first two alphabets then it will become "ice", which is really cold.

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.