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

Revision as of 17:22, 23 August 2021

Problem 2

A softball team played ten games, scoring $1,2,3,4,5,6,7,8,9$ , and $10$ runs. They lost by one run in exactly five games. In each of the other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?

$\textbf {(A) } 35 \qquad \textbf {(B) } 40 \qquad \textbf {(C) } 45 \qquad \textbf {(D) } 50 \qquad \textbf {(E) } 55$

Solution

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 $(2+4+6+8+10)+(1+2+3+4+5) = 55$ , which is $E$

Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=91 (problem 2 starts at 1:31 in the video)

~sugar_rush

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 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 <math>(2+4+6+8+10)+(1+2+3+4+5) = 45</math>, which is <math>C</math>
+Therefore, the total runs by the opponent is <math>(2+4+6+8+10)+(1+2+3+4+5) = 55</math>, which is <math>E</math>
 ==Video Solution==
 https://www.youtube.com/watch?v=2vf843cvVzo?t=91 (problem 2 starts at 1:31 in the video)

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

Revision as of 17:22, 23 August 2021

Contents

Problem 2

Solution

Video Solution

See also