2013 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2013 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10A solutions, or check the answer key.

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Concepts:paritycasework

Difficulty rating: 1070

4.

A softball team played ten games, scoring 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, 7,7, 8,8, 9,9, and 1010 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?

3535

4040

4545

5050

5555

Solution:

Note that if they scored twice as many runs as their opponents, then they scored an even number of runs.

This means in the games where they scored 2,4,6,8,2, 4, 6, 8, and 1010 runs, their opponents scored 1,2,3,4,1, 2, 3, 4, and 55 runs respectively.

This sums to 1+2+3+4+5=15. 1 + 2 + 3 + 4 + 5 = 15.

In the other games, their opponents scored 2,4,6,8,2, 4, 6, 8, and 1010 runs.

This sums to 2+4+6+8+10=30. 2 + 4 + 6 + 8 + 10 = 30.

The total number of runs is then 15+30=45. 15 + 30 = 45. Thus, C is the correct answer.

Problem 4 in Other Years