ANSWER- 2hr 11min

EXPLANATION-
This question cannot be solved unless you know a fact i.e. the hour hand is exactly on the minute mark five times every hour – on the hour, 12 minutes past hour, 24 minutes past hour, 36 minutes past hour and 48 minutes past hour.

Denoting the number of hours as X and the number of minutes as Y, and lets begin with the calculation.

When we have the hour hand on the minute mark,
The position of hour hand: 5X + Y/12
And the position of the minute hand = Y

When the first observation was taken,
Y = Y = 5X + Y/12 + 6
Or, 60X = 11Y – 72.

With the given facts, we know for sure that Y can only be 0, 12, 24, 36 or 48.
According to which, the possible values for X and Y can be 1 and 12 respectively. Thus the time must be 1:12 when the first observation was taken.

Similarly, in the second observation, let’s make the equation:
60X = 11Y – 84
Here, the possible values for X and Y here are 3 and 24 respectively. According to this, the time is 3:24 for sure.

From 1:12 to 3:24, two hours and eleven minutes have passed.

ANSWER- 6h27m41.5s

EXPLANATION-
You need to solve (n+(180−θ)/360)×30=θ, where n is the number of hours past 6 o’clock. The solution is
θ=18013(2n+1)deg

For different values of θ you would get different times.
For example, for n=6 you get θ=180 which is 12 o’clock.
For n=0 you get θ=180/13=13.8, which, using 30 degree=5 min, is  6h27m41.5s

The longest line of friends that can have chocolate is 13, with the last person taking all the remaining chocolates.
There are not enough chocolates for a fourteenth person, because they would have to take 14 chocolates (after the previous person had taken 13) and there would be only 9 left.

He is the man’s son.

EXPLANATION-
Put YOURSELF in the riddle.

The riddle tells us, “that man’s father…” IS MY FATHER’S SON. Thus, the the father of the man in the picture is YOUR father’s son. In other words, YOU. That establishes the relationship between the FATHER of the man in the picture, and YOU.

So, WHO is in the picture? It can ONLY be your SON. So, the father of your son (who is the one pictured) is your father’s son.

Answer – 4

EXPLANATION-
count is giving the number of circles in the number
8080 = 6 (six circles)
1357 = 0
2022 = 1
1999 = 3
6666 = 4 (Four circles)

The ball will make infinite bounces before it is stopped.
As per the puzzle it will keep on bouncing half way up every time it hits the ground. But gravity will play its part at certain point of time which will make it stop.

ANSWER – 0.3611

EXPLANATION-
The probability of either the Rook or the Bishop attacking the other is 0.3611

A Rook and a Bishop on a standard chess board can be arranged in:
= 64P2  (64 Permutation 2)
= 64*63
= 4032 ways

Now,
There are 2 possible cases – Rook attacking Bishop and Bishop attacking Rook.

As we know that the Rook and the Bishop never attack each other simultaneously. Let’s consider both the cases one by one.

Case I – Rook attacking Bishop
Rook can be placed in any of the given 64 positions and it always attacks 14 positions.
Hence, total possible ways of the Rook attacking the Bishop
= 64*14
= 896 ways

Case II – Bishop attacking Rook
View the chess-board as a 4 co-centric hollow squares with the outermost square with side 8 units and the innermost square with side 2 units.

If the bishop is in one of the outer 28 squares, then it can attack 7 positions.
If the bishop is in one of the 20 squares at next inner-level, then it can attack 9 positions.
If the bishop is in one of the 12 squares at next inner-level, then it can attack 11 positions.
If the bishop is in one of the 4 squares at next inner-level (the innermost level), then it can attack 13 positions.

Hence, total possible ways of the Bishop attacking the Rook
= 28*7 + 20*9 + 12*11 + 4*13
= 560 ways
Thus, the required probability is
= (896 + 560) / 4032
= 13/36
= 0.3611

This problem is intriguing!

Number the sentences in conversation:
<<1>> B: I can’t tell what the numbers are.
<<2>> A: I can’t tell what the numbers are.
<<3>> B: I can’t tell what the numbers are.
<<4>> A: I can’t tell what the numbers are.
<<5>> B: I can’t tell what the numbers are.
<<6>> A: I can’t tell what the numbers are.
<<7>> B: Now I can tell what the numbers are.

Denote the sum S and sum of squares N. S-list will be a list of all possible sums a + b, N-list a list of all possible sums of squares a^2 + b^2. Initially, the S-list contains all positive integers >1, N-list all positive integers expressible as sum of two positive integer squares.

The section <<i>> below describes the information gained after the line <<i>> in the conversation was spoken.

<<1>> N is not expressible as the sum of positive integer squares uniquely. N-list is correspondingly adjusted (numbers expressible as sum of two positive integer squares uniquely are thrown out). S-list is adjusted – sums are thrown out, that don’t allow for summands a, b, such that a^2 + b^2 is on the current N-list.

<<2>> S allows for more values of a, b (a + b = S), such that a^2 + b^2 belongs to the N-list. S-list is adjusted (sums are deleted not allowing for more values of a, b (a + b = the sum), such that a^2 + b^2 belongs to the N-list).

<<3>> N allows for more values of a, b (a^2 + b^2 = N), such that a + b is on the S-list. Those integers that don’t are deleted from the N-list.

<<4>> S allows for more values of a, b, such that a^2 + b^2 belongs to the N-list. Those integers that don’t are deleted from the S-list.

<<5>> N allows for more values of a, b, such that a + b belongs to the S-list. Those integers that don’t are deleted from the N-list.

<<6>> S allows for more values of a, b, such that a^2 + b^2 belongs to the N-list. Those integers that don’t are deleted from the S-list.

<<7>> At this moment there is only one pair of positive integers a, b, such that a + b is on the S-list and a^2 + b^2 = N. If N-list contains more values, they need to be checked to see if they allow for unique solution.

If some bounds are imposed on a,b, I guess the solution could be found by computer or manual deleting from the S- and N-list. I haven’t really thought beyond this outline yet, at this moment solving the problem in full generality seems difficult… I’d love to see the intended solution.

SOURCE– http://www.artofproblemsolving.com/community/c146h150971