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:

There is no reason whatever why the customer's original deposit of Rs.100 should equal the total of the balances left after each withdrawal.
The total of withdrawals in the left-hand colum may equal Rs.100, but is is purely coincidence that the total of the right-hand column is close to Rs.100.
Let us show another example, but starting with Rs.200 in the bank:
Withdrawals Balance left
Rs.50 Rs.150
Rs.25 Rs.125
Rs.10 Rs.115
Rs.8 Rs.107
Rs.5 Rs.102
Rs.2 Rs.100
Rs.100 Rs.699
Moral of this story? Don't Total Balances

Solution:

Solution is shown in image below

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

Playing chess

Game needs two players so Brother-8 is playing chess with Brother-2

Solution:

I filled a glass full of water and held it over my head for 20 minutes.

Solution:

28 =>1+2+4+7+14

cannot be 1 (as number cannot be included)

Solution:

We now have a square of 3x3 size, 2 squares of 2x2 size and 5 squares of 1x1 size.

Solution:

40 balls

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

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.