Difference between revisions of "2013 AMC 10A Problems/Problem 4"
Countingkg (talk | contribs) |
Countingkg (talk | contribs) |
||
| Line 28: | Line 28: | ||
\end{tabular}</math> | \end{tabular}</math> | ||
| − | The sum of their opponent's scores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = | + | The sum of their opponent's scores is <math>2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{\textbf{(C) }45}</math> |
==See Also== | ==See Also== | ||
Revision as of 14:21, 8 February 2013
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?
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:
The sum of their opponent's scores is 
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 | |
