Difference between revisions of "2013 AMC 12A Problems/Problem 2"

(Created page with "To score twice as many runs as their opponent, the softball team must have scored an even number. Therefore we can deduce that when they scored 1, 3, 5, 7, and 9 runs, they lost...")
 
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To score twice as many runs as their opponent, the softball team must have scored an even number.
 
To score twice as many runs as their opponent, the softball team must have scored an even number.
  
Therefore we can deduce that when they scored 1, 3, 5, 7, and 9 runs, they lost by one run, and
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Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much.
when they scored 2, 4, 6, 8, and 10 runs, they scored twice as many runs as their opponent.
 
  
Therefore, the total runs by the opponent is (2+4+6+8+10)+(1+2+3+4+5), which is 45, or C
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Therefore, the total runs by the opponent is <math>(2+4+6+8+10)+(1+2+3+4+5) = 45</math>

Revision as of 04:20, 7 February 2013

To score twice as many runs as their opponent, the softball team must have scored an even number.

Therefore we can deduce that when they scored an odd number of runs, they lost by one, and when they scored an even number of runs, they won by twice as much.

Therefore, the total runs by the opponent is $(2+4+6+8+10)+(1+2+3+4+5) = 45$

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