Difference between revisions of "2013 AMC 12A Problems/Problem 2"
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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>, which is <math>C</math> | 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> | ||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=2vf843cvVzo?t=91 (problem 2 starts at 1:31 in the video) | ||
+ | |||
+ | ~sugar_rush | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2013|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:08, 23 November 2020
Contents
Problem 2
A softball team played ten games, scoring , and 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?
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 , which is
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=91 (problem 2 starts at 1:31 in the video)
~sugar_rush
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.