Solution:

24

Explanation:
21 * 4 - (4 * 4) = 68
12 * 4 - (4 * 4) = 32
17 * 4 - (4 * 4) = 52
10 * 4 - (4 * 4) = 24

Solution:

Tilt the barrel until the fruit juice barely touches the lip of the barrel. If the bottom of the barrel is visible then it is less than half full.If the barrel bottom is still completely covered by the fruit juice, then it is more than half full.

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:

The equation can be solved by substituting
B = 7
A = 8
S = 8
E = 3
L = 5

Therefore the equation now looks as:
7483 +
7455
-------
14938
-------

Solution:

If we have to measure precisely, it is impossible. Because we are asked to measure odd number of liters whereas all the jars we have can contain only even liters of water.

* Tricky question asked in ias/infosys/infoedge/ibibo interview

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.

Solution:

1653

Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653

Solution:

1/36

Explanation:
The number of possibilities for the first as well as the second dice are 6 and 6 respectively. Therefore the total possibilities or outcome are 6 * 6 = 36.
Out of 36 outcomes, we need only one case i.e. the first gives 2 and the second gives 5.
Therefore, the probability will be 1 on 36 or 1/36.

Solution:

Ganpat had a black suit and a brown suit.
Harish had a brown suit and a pink suit.
Manoj had a white suit and a black suit.
Tarun had a pink suit and a white suit.

Solution:

If you ask the guard who always tells the truth, he would tell you the other guard would point you to the door of death. If you ask the guard who always lies, he would tell you the opposite door of the truth-telling guard and point you to the door of death. In either case, both guards will point to the door of death so you should choose the other one.

Solution:

Solution is below