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)
- B)
- C)
- D)
- E)
Answer
C
Key insight
A game won by scoring twice the opponent's runs needs an even score, so the wins are the even games and the one-run losses are the odd games.
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: . The opponents scored
runs in those games.
The remaining five games, with scores , were lost by exactly one run, so the opponents scored one more each:
Total opponent runs: . The answer is .
Why this works
The problem never says which games were won, so a hidden integrality condition must decide it: "twice the opponent" forces an even team score, and there are exactly five even scores. Whenever a word problem seems under-determined, look for a divisibility or parity constraint that pins down the missing assignment.
The trap
Assuming the five wins were the five highest-scoring games (6 through 10), which double-counts and gives a wrong total.
Common mistakes
- Assuming the five wins were the five highest-scoring games (6 through 10), which double-counts and gives a wrong total.
- Computing the opponents' runs in the losses as one less than the team's score instead of one more.
Techniques
Set up the equation/formula and compute; no special trick needed · Use an invariant, parity, or coloring argument