Solution:

No wheels which imply car is not moving :-)

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.

Solution:

Everyone will done one day or other.

Abraham Lincoln era is too old and we expect most of them(if not all) to be dead by now.

Solution:

At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple

Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.

Solution:

The Lady sitting in the chair has the detector attached to the back of her blouse.

Solution:

Move the coins as given in the figure and you will get the required result.

Solution:

Colors:
Y B R G
G R B Y
B Y G R
R G Y B


Signs:
x + - /
- / x +
/ - + x
+ x / -

Solution:

7 Races.

Solution:

It will take the man 58 minutes to reach the top.

Since he is climbing 3 feet and falling 2 feet back every passing minute, he is actually climbing 1 feet every minute.
Therefore, he would have climbed 57 feet in 57 minutes. The remaining distance is 3 feet. He will take one more minute to climb it and reach the surface. Since he has reached the top surface, he won't slip back.

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.