Solution:

9376

Looking at the last digit, the last digit must be either 0, 1, 5 or 6.
Then looking at the last two digits, the last two digits must be either 00, 01, 25 or 76.
Then looking at the last three digits, the last three digits must be either 000, 001, 625 or 376.
Then looking at the last four digits, the last four digits must be either 0000, 0001, 0625 or 9376.
Out of those, only 9376 is a 4 digit number

Solution:

It was only a trick

I borrowed my parents (Pound 50 + Pound 50 = Pound 100).
I bought the book for Pound 97 and i had Pound 3 change.
So, the money that i borrowed and bought the book are Pound 100 (borrowing money)
and the change Pound 3 also the borrowing money the book is Pound 97
Total: Pound 97 + Pound 3(the change) = Pound 100
The money that i used to buy the book all i borrowed.
Pound 97 is the borrowing money, change: Pound 3 also the borrowing money. Pound 97 + Pound 3 = Pound 100

Solution:

== Mathematical Answer ==
Mathematical Answer = number of seconds in a minute * number of minute in an hour * number hour in a day * number of day in year
60 * 60 * 24 * 365 = 31536000
60 * 60 * 24 * 365.25 = 31557600 (if you want account for leap year)


== We are looking for trick answer (which is 24) ==
Every month contains only 2 second like
A) 2nd January ("second" January)
B) 22nd January (twenty"second" January)

Therefore 2 * 12 months = 24.

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:

Move and then pour all juice from second glass to fifth glass.

Solution:

White should move the rook from
A1 to D1 and hence complete the castling move.

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

Solution:

What is the correct time ?

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. Fill 5 liters jar ( 5l-jar:5, 3l-jar:0)
2. Transfer to 3 liters pail (5l-jar:2, 3l-jar:3)
3. Empty 3 liters jar ( 5l-jar:2, 3l-jar:0)
4. Transfer 2q from 5 pail to 3 pail (5l-jar:0, 3l-jar:2)
5. Fill 5 liters pail(5l-jar:5, 3l-jar:2)
6. Transfer 1q from 5 pail to 3 pail(5l-jar:4, 3l-jar:3)

Solution:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.