If u know the rules of cricket here is a Q to test ur method of approach?
Seven runs to win, two batsmen at the crease, both at 94 runs each. Two balls are remaining. The team won & both batsmen made centuries. How could it be possible?
Answers:
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First batsman takes a single, the ball goes behind him, hits the wicket keeper's helment for which he's awarded 5 runs...he makes his 100 there.Second batsman is now on strike, with one ball to go, he just hits a 6 and the game is over.
Nothing else comes to mind right now. That's the only way I know that you can get 6 runs and get off strike at the same time.
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The striker hits a six of the last ball of the over,in the next over the other bats man of yours hits a six and answers your question.What are your plans for the world cup 2007? Who will take the trophy this year?
I think "riskbreaker" has the only correct answer.Who wants to support pakistan cricket website plz do?
hhhm. gud question...okkk.. see this is wat haapenssss.... first the first ball is done....lets imagine it is kevin there... hmm...it is kinda confusingg
hmm he hits a four and takes a run... I DONT KNOWWW. :(
soorrieee. plllz tell mee... is the first guy ryte??
The first batsman, who is on 94 runs hit the ball and run for a run. The ball hits on the helmet kept behind the wicket keeper before it is fielded. Thus the batsman gets 6 runs (1 run he ran and 5 extra runs for the ball hitting the helmet) and
completes century. Next ball is faced by other batsman who has crossed over. He hit the ball for a six and completes his century also and the team also wins.
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1st batsman hit a 6 at da last ball of the over.2nd batsman hit a 6 again at da 1st ball of da next over.simple.y r u guys writin so much?r u all dumb or wht?:-/I can only think of one way that this is possible. I note that you do not specify how many wickets have fallen. I'm assuming less than nine have fallen.
Batsman #1 on strike hits the ball towards the boundary and the batsman run 3 runs and continue for a fourth run. The ball is cutoff by a fielder who throws in wildly, and the ball is overthrown. The runners complete the fourth run and continue running completing 2 more runs (6 runs) and continuing for an additional run. A fielder fields the overthrow and throws in knocking the stumps of the non-strikers end before the batsmen complete the seventh run. Batsman #1 has scored six (completing his century) but is runout going for the seventh run.
A new batsman comes in for the last ball, but he is at the non-strikers end. Batsmen #2 is now at the strikers end to face the last ball needing only one run to win, and he hits a six, completing his century and winning the game.
UPDATE: It also occurs to me that in the scenario above, the runners can complete the seventh run, be judged to have run 'one short' so that only six runs count. One run will still be required, but the batsman will have rotated the strike. In this case, it wouldn't matter if it was already 9 wickets down.
Riskbreaker & Vakayil, you're idea is good, but I believe if the ball hit the helmet behind the wicket keeper the batting team would receive 5 PENALTY runs, which would not be attributed to the batters total.
Mohammed & ghost, the question implies that there are two balls left in the final over. There is not another over available for the batsmen to rotate the strike.
Lara, I'm not positive but I think even if the batsmen cross once when the ball is hit in the air, the batsmen that hit the six retains the strike when the ball is hit over the boundary. If he retires, a new batsman would be on strike, not the second batsman on 94.
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