AMC 10 · 2013 · #4
Easy mode Grade 3A softball team played ten games. Their scores were 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs, one score in each game. In exactly five games they lost by just one run. In each of the other five games they scored twice as many runs as the other team. How many runs did the other teams score altogether?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A team's ten games have scores $1,2,3,\dots,10$. In five games the team lost by exactly one run; in each of the other five the team scored twice its opponent. Add up all the opponents' runs.
Givens: The team's ten scores are $1,2,3,4,5,6,7,8,9,10$, one per game; In exactly five games the team lost by one run (opponent scored one more); In each of the other five games the team scored twice as many runs as its opponent; Answer choices: (A) $35$, (B) $40$, (C) $45$, (D) $50$, (E) $55$
Unknowns: The total number of runs scored by all the opponents across the ten games
Understand
Restated: A team's ten games have scores $1,2,3,\dots,10$. In five games the team lost by exactly one run; in each of the other five the team scored twice its opponent. Add up all the opponents' runs.
Givens: The team's ten scores are $1,2,3,4,5,6,7,8,9,10$, one per game; In exactly five games the team lost by one run (opponent scored one more); In each of the other five games the team scored twice as many runs as its opponent; Answer choices: (A) $35$, (B) $40$, (C) $45$, (D) $50$, (E) $55$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #13 Convert to Algebra, #7 Identify Subproblems
The hard part is deciding which five scores are losing games and which five are doubling games. Tool #13 (Convert to Algebra) turns the two rules into number relationships. Then Tool #3 (Eliminate Possibilities) cracks it with one parity fact: a doubling score is $2\times$ a whole number, so it must be even — which forces the five odd scores to be the losing games and the five even scores to be the doubling games. Tool #7 (Identify Subproblems) splits the count into two easy sums that are added at the end.
Execute — Answer: C
2.OA.A.1 Step 1 Turn each rule into a number relationship
- Write what each kind of game means for the opponent's score.
- In a doubling game the team scored twice the opponent, so the team's score is $2\times(\text{opponent})$, which means the opponent scored half the team's score.
- In a losing game the team lost by one run, so the opponent scored one more than the team, that is $(\text{team})+1$.
💡 Rewriting the words as "half of" and "one more than" tells you exactly how to get each opponent's score from the team's score.
2.OA.C.3 Step 2 Use odd vs even to sort the games
- A doubling score equals $2\times$ a whole number, so it must be even.
- The odd scores $1,3,5,7,9$ can never be twice a whole number, so those five games cannot be doubling games — they must be the five losing games.
- That leaves the even scores $2,4,6,8,10$ as the five doubling games.
- The counts match perfectly: five and five.
💡 Doubling always lands on an even number, so any odd score is stuck being a losing game.
2.NBT.B.5 Step 3 Add the opponents in the losing games
- For the odd scores the opponent scored one more than the team.
- So the opponents scored $1{+}1,\;3{+}1,\;5{+}1,\;7{+}1,\;9{+}1$, which are $2,4,6,8,10$.
- Adding these gives the opponents' runs from the losing games.
💡 Each losing opponent is just the team's score bumped up by one.
3.OA.A.3 Step 4 Add the opponents in the doubling games
- For the even scores the opponent scored half the team's score.
- So the opponents scored $\tfrac{2}{2},\;\tfrac{4}{2},\;\tfrac{6}{2},\;\tfrac{8}{2},\;\tfrac{10}{2}$, which are $1,2,3,4,5$.
- Adding these gives the opponents' runs from the doubling games.
💡 Each doubling opponent is just the team's score cut in half.
2.NBT.B.5 Step 5 Combine the two totals
- Add the runs from both kinds of games: $30$ from the losing games and $15$ from the doubling games.
- The opponents scored $30+15=45$ runs in all, which is choice (C).
💡 The two groups cover all ten games with no overlap, so their totals simply add.
2.OA.A.1 Write what each kind of game means for the opponent's score. In a doubling game 2.OA.C.3 A doubling score equals $2\times$ a whole number, so it must be even. The odd sc 2.NBT.B.5 For the odd scores the opponent scored one more than the team. So the opponents 3.OA.A.3 For the even scores the opponent scored half the team's score. So the opponents 2.NBT.B.5 Add the runs from both kinds of games: $30$ from the losing games and $15$ from Review
Reasonableness: The team scored $1+2+\cdots+10=55$ runs. In losing games opponents scored a bit more than the team ($+1$ each, five games $=+5$), and in doubling games opponents scored far less (half each). A total of $45$ sits sensibly below the team's $55$, matching (C). A quick sanity split confirms it: losing opponents $30$ against team's odd total $25$ (five more, as expected), doubling opponents $15$ against team's even total $30$ (exactly half).
Alternative: Skip the split and use the halves-and-plus-one structure directly. Odd team scores sum to $25$, and each losing opponent adds one more, giving $25+5=30$. Even team scores sum to $30$, and each doubling opponent is half, giving $\tfrac{30}{2}=15$. Total $30+15=45$, again (C).
CCSS standards used (min grade 3)
2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Translating "lost by one run" into opponent $=$ team $+1$.)2.OA.C.3Determine whether a group of objects has an odd or even number (Recognizing that a doubling score must be even, so the odd scores must be the losing games.)3.OA.A.3Solve multiplication and division word problems within 100 (Halving each even team score to get the doubling opponents $1,2,3,4,5$.)2.NBT.B.5Fluently add and subtract within 100 (Summing $2+4+6+8+10=30$, $1+2+3+4+5=15$, and $30+15=45$.)
⭐ Doubling always makes an even number, so the odd scores had to be the close losses — sort the games that way, then add the halves and the plus-ones.
⭐ Doubling always makes an even number, so the odd scores had to be the close losses — sort the games that way, then add the halves and the plus-ones.
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