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Difference between revisions of "2013 AMC 10A Problems/Problem 4"
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The sum of their opponent's scores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}</math>
 
The sum of their opponent's scores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}</math>
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/Z6SSR22FIWg
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~Education, the Study of Everything
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==Video Solution==
 
==Video Solution==

Revision as of 12:09, 1 July 2023

Problem

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 their 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

We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by $2$. Thus, from this, we know that the five games where they lost by one run were when they scored $1$, $3$, $5$, $7$, and $9$ runs, and the others are where they scored twice as many runs. We can make the following chart:

$\begin{tabular}{|l|l|} \hline Them & Opponent \\ \hline 1 & 2 \\ 2 & 1 \\ 3 & 4 \\ 4 & 2 \\ 5 & 6 \\ 6 & 3 \\ 7 & 8 \\ 8 & 4 \\ 9 & 10 \\ 10 & 5 \\ \hline \end{tabular}$

The sum of their opponent's scores is $2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}$

Video Solution (CREATIVE THINKING)

https://youtu.be/Z6SSR22FIWg

~Education, the Study of Everything


Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=91

~sugar_rush

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 10 Problems and Solutions
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

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