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2013 AMC 10A Problem 4

2013 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 2 min · position P1-10

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?

A) $35$
B) $40$
C) $45$
D) $50$
E) $55$

Answer

Key insight

A win means the team scored double the other team, so its score must be even. That makes the even scores the wins and the odd scores the one-run losses.

Solution

The problem does not say which games were won. So first figure that out.

Which games were wins? In a win, the team scored twice as many runs as the other team. Twice a whole number is always even. So a win must have an even score: $2, 4, 6, 8$ or $10$ .
That is exactly five scores, and there are exactly five wins. So the wins are the even scores, and the five losses are the odd scores $1, 3, 5, 7, 9$ .
Opponents in the wins. The opponent scored half of the team's score:

Team scored	2	4	6	8	10
Opponent scored	1	2	3	4	5

That is $1 + 2 + 3 + 4 + 5 = 15$ runs.

Opponents in the losses. Losing by one run means the opponent scored one more than the team:

Team scored	1	3	5	7	9
Opponent scored	2	4	6	8	10

That is $2 + 4 + 6 + 8 + 10 = 30$ runs.

Total. $15 + 30 = 45$ .

The answer is $\boxed{\textbf{(C)}\ 45}$ .

Why this works

When a word problem seems to leave something out (here: which games were won), look for a hidden rule that decides it. "Twice as many" forces an even number, and that fixed which games were wins. Even/odd reasoning like this often fills in the missing piece.

The trap

Guessing that the wins were the five biggest scores, 6 through 10. Scores like 7 and 9 cannot be double a whole number.

Concise solution

In a game the team won, they scored twice the opponent's runs, so the opponent scored half of the team's score. Runs are whole numbers, so those five games must be the ones with even scores: $2, 4, 6, 8, 10$ . The opponents scored
$1 + 2 + 3 + 4 + 5 = 15$
runs in those games.

The remaining five games, with scores $1, 3, 5, 7, 9$ , were lost by exactly one run, so the opponents scored one more each:
$2 + 4 + 6 + 8 + 10 = 30.$

Total opponent runs: $15 + 30 = 45$ . The answer is $\boxed{\textbf{(C)}\ 45}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)

Techniques

Set up the equation/formula and compute; no special trick needed · Use an invariant, parity, or coloring argument