24 October 2008
2 BATSMEN EACH ON 94, 7 RUNS TO WIN IN 3 BALLS ,BOTH MAKE UNBEATEN 100.HOW???THIS QUESTION WAS ASKED IN AN INFOSYS INTERVIEW!! CAN U CRACK IT????
25 October 2008
Mr.Jitendra, it is not possible as you said, because:
1)when the first man hits six, he continues to bat, second man will not get the second ball.
2)Even if the second man gets a chance and hits a four in the second ball, the match will be over as the required runs are scored and therefore the second man will not get the third ball to complete his 100.
25 October 2008
1st bALL : ONE BATSMAN HIT A FOUR.
2nd Ball : same batsman make a shot and take three runs but onje run is short and it will be counted as 2 runs. 3rd Ball : the other batsman hit a SIX and won the match.
Both are at unbeaten 100.
Also the case is the overs might be reduced to 48.2 overs or 48.1 overs something then both batsman can hit a SIX
25 October 2008
But in infosys interview, the answer was like this: Case 1: A batsman can be given out
1st batsman hits a six....gets caught on d nxt ball...crease is changed....next batsman hits a six again...
Case 2: No batsman is out
1st batsman hits d ball n hits d keeper'ds helmet kept behind...he also takes a single...6 runs are added to his total making it 100...on d next ball ,2nd batsman hits a six,making his score 100....as simpl as dat....
CAclubindia
