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There is an 8 by 8 chessboard in which two diagonally opposite corners
have been cut off.You are given 31 dominos, and a single domino can
cover exactly two squares. Can you use the 31 dominos to cover the
entire board?
Simple Arrangement? Check out it's possibility!
What was the challenge given?
Initial mathematical calculations might suggest that the task is pretty simple. If 2 square are cut off from 64 squares then 62 squares will be left which are enough for 31 dominos (each covering 2 squares).
But, that is not the case. Since, 2 diagonally opposite squares are removed, they has to be either black or white like shown below with shaded regions.
We need 1 black and 1 White square for placement of 1 domino on the chessboard.That is 31 Black and 31 White squares are needed to give cover for 31 dominos.
In the above 2 cases, there are either 32 Black and 30 White or 30 Black and 32 White squares are available.
This makes the task of placing 31 dominos on the chessboard (whose 2 diagonally opposite squares are removed) impossible!
Three friends (A, B and C) are playing ping pong. They play the
usual way: the winner stays on, and the loser waits his/her turn again.
At the end of the day, they summarize the number of games that each of
them played:
A played 10
B played 15
C played 17.
Who lost the second game?
This person played & lost the second game!
How games were played?
A played 10, B played 15 and C played 17 games. So total number of presences are 10 + 15 + 17 = 42. Every 2 presences form a game. Hence, the number of games played are 42/2 = 21.
Let's take into consideration the minimum number of games that a player can play. For that, he need to loose every game that he has played. That is, if he has played first game then he must have out in second but replaced looser of second in third game. In short, he must have played odd numbered of games like 1,3,5,7,9,11,13,15,17,19,21.That's 11 games in total.
And if he had made debut in second game then he must had played even numbered games like 2,4,6,8,10,12,14,16,18,20. That's 10 games in total.
Since, in the case only A has played 10 games, he must have made debut in second game where he lost that game to make comeback in 4th game thereby replacing looser of third game.
In a one day international cricket match, considering no extras(no wides, no ‘no’ balls, etc.) and no overthrows.
What is the maximum number of runs that a batsman can score in an ideal case ?
Note: Here we assume ideal and little practical scenario. We assume
that batsman can not run for more than 3 runs in a ball, as otherwise
there is no limit, he can run infinite runs(theoretically) in a ball, as
far as opposite team does not catch the ball.”
Could be tricky! Here is correct number!
What was the question?
It's not as straight forward as it seems at first glance. That is one might think that the maximum score that one can score by hitting 6 on every ball of 50 overs is 50 x 6 x 6 = 1800.
No doubt, 1800 can be the maximum team score but not the individual score.Since batsman rotates strike every over, here both batsmen share these 1800 runs as 900 to each.
However, if the batsman on strike runs 3 runs on the last ball of the over then he can hit 5 more sixes in next over as strike is rotated back to him in next over. He can continue in this way till 49th over. And in 50th over he can hit 6 sixes on 6 balls.
Maximum Individual Score = 49 x [(5x6)+3] + 36 = 1617 + 36 = 1653.
In this case, the batman at the non-striker end scores 0 runs as he doesn't get strike on a single ball.
King Octopus has servants with six, seven, or eight legs. The
servants with seven legs always lie, but the servants with either six or
eight legs always say the truth.
One day, 4 servants met :
The blue one says: “Altogether we have 28 legs”;
the green one says: “Altogether we have 27 legs”;
the yellow one says: “Altogether we have 26 legs”;
the red one says: “Altogether we have 25 legs”.
What is the color of the servant that says the truth?
"I'm telling the truth and my color is..."
