AMC 10 · 2013 · #2
Easy mode Grade 4A softball team played ten games. Their own scores were 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs — one of those scores in each game. In exactly five of the games they lost by one run. In each of the other five games they scored exactly twice as many runs as the other team. How many runs did the other teams score in total?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A softball team played ten games. Their own scores in those games were $1,2,3,4,5,6,7,8,9,10$ — each of those totals used in exactly one game. In exactly five of the games they lost by exactly one run. In each of the other five games their score was exactly double their opponent's score. Find the total number of runs the opponents scored across all ten games.
Givens: Ten games were played, and the team's scores were $1,2,3,4,5,6,7,8,9,10$, each value used exactly once; In exactly $5$ games the team lost by one run, so in each of those the opponent scored one more than the team; In each of the other $5$ games the team scored twice as many runs as the opponent; Answer choices: (A) $35$, (B) $40$, (C) $45$, (D) $50$, (E) $55$
Unknowns: The total number of runs scored by the opponents over all ten games
Understand
Restated: A softball team played ten games. Their own scores in those games were $1,2,3,4,5,6,7,8,9,10$ — each of those totals used in exactly one game. In exactly five of the games they lost by exactly one run. In each of the other five games their score was exactly double their opponent's score. Find the total number of runs the opponents scored across all ten games.
Givens: Ten games were played, and the team's scores were $1,2,3,4,5,6,7,8,9,10$, each value used exactly once; In exactly $5$ games the team lost by one run, so in each of those the opponent scored one more than the team; In each of the other $5$ games the team scored twice as many runs as the opponent; Answer choices: (A) $35$, (B) $40$, (C) $45$, (D) $50$, (E) $55$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
The opponent's score in a game is easy to compute once you know which type of game it was, so Tool #7 (Identify Subproblems) splits the opponent total into two group totals: the five one-run losses and the five doubling games. The real work is that the problem never says which scores land in which group, and that is where Tool #3 (Eliminate Possibilities) is primary. Halving an odd score would give a fraction of a run, so odd scores are eliminated from the doubling group; then counting shows five doubling games have only five even scores available, which forces the split completely. Tool #4 (Introduce a Variable) is held in reserve for the review: naming the loss-game score total lets the whole answer be written as one expression, a check that runs on different machinery.
Execute — Answer: C
4.OA.A.1 Step 1 Two game types, two opponent formulas
- Handle the two kinds of games separately.
- If the team scored $s$ runs and lost by one, the opponent scored $s + 1$.
- If the team scored $s$ runs and that was twice the opponent's, then $s = 2o$, so the opponent scored half of $s$.
- Every game is exactly one of these two types, five games each.
- So the opponent total is the sum of $s+1$ over the five losses plus the sum of $\tfrac{s}{2}$ over the five doubling games.
- The only missing piece is which five of the scores $1$ through $10$ go into each group.
💡 "Twice as many" is a multiplication comparison, so reading it backwards turns the team's score into the opponent's — just halve it.
4.OA.B.4 Step 2 A doubling game needs an even score
- In a doubling game the opponent scored $\tfrac{s}{2}$ runs.
- There is no such thing as half a run, so $\tfrac{s}{2}$ must be a whole number, which means $s$ is a multiple of $2$.
- Among $1$ through $10$ the multiples of $2$ are exactly $2, 4, 6, 8, 10$.
- So none of the odd scores $1, 3, 5, 7, 9$ can come from a doubling game; each of those five games must be one of the one-run losses.
💡 An odd number of runs cannot be split into two equal whole halves, so an odd score can never be double a whole score.
4.OA.A.3 Step 3 Five evens fill five slots exactly
- Ruling odd scores out of the doubling group is only half the argument.
- On its own it leaves room for an even score such as $4$ to have been a one-run loss too.
- Count instead.
- The doubling group contains exactly five games, their five scores are all different, and every one of them must come from the five-number set $\{2,4,6,8,10\}$.
- Choosing five different values out of a set that has only five values leaves no freedom at all: the doubling scores are exactly $2,4,6,8,10$, and the one-run losses are exactly $1,3,5,7,9$.
- This split is also genuinely possible — the opponents score $2,4,6,8,10$ in the losses and $1,2,3,4,5$ in the doubling games, all whole numbers — so the situation described really exists, and this is its only version.
💡 When five different items must fit into exactly five boxes, every box gets filled — there is no other arrangement.
4.NBT.B.4 Step 4 Add the two group totals
- In the five one-run losses the team scored $1,3,5,7,9$, so the opponents scored $2,4,6,8,10$, which totals $30$.
- In the five doubling games the team scored $2,4,6,8,10$, so the opponents scored half of each, namely $1,2,3,4,5$, which totals $15$.
- Adding the two groups gives $30 + 15 = 45$ opponent runs, so the answer is $\textbf{(C)}\ 45$.
💡 Once every game's opponent score is pinned down, the total is just two short sums added together.
4.OA.A.1 Handle the two kinds of games separately. If the team scored $s$ runs and lost b 4.OA.B.4 In a doubling game the opponent scored $\tfrac{s}{2}$ runs. There is no such thi 4.OA.A.3 Ruling odd scores out of the doubling group is only half the argument. On its ow 4.NBT.B.4 In the five one-run losses the team scored $1,3,5,7,9$, so the opponents scored Review
Reasonableness: Write out all ten games and check them one at a time. The losses are $1$–$2$, $3$–$4$, $5$–$6$, $7$–$8$, $9$–$10$: each is a loss by exactly one run. The wins are $2$–$1$, $4$–$2$, $6$–$3$, $8$–$4$, $10$–$5$: in each the team's score is exactly double. That is five of each type, the team's scores are exactly $1$ through $10$ with no repeats, and the opponent runs add to $2+4+6+8+10+1+2+3+4+5 = 45$. The size is sensible too: the team scored $55$ runs in all, and a team that wins five games by doubling and loses five by a single run should give up a bit less than it scores. A rough bound confirms the scale: whatever the split, the opponent total is $\tfrac{T+65}{2}$ where $T$ is the team's score total in the loss games, and $T$ ranges between $1+2+3+4+5 = 15$ and $6+7+8+9+10 = 40$, so the opponent total must lie between $40$ and $52.5$. That alone rules out (A) $35$ and (E) $55$ before any parity reasoning, and $45$ sits comfortably inside.
Alternative: Avoid sorting the games one by one and work with totals instead. Let $T$ be the total of the team's scores in the five one-run losses; the five doubling scores then total $55 - T$, since $1 + 2 + \cdots + 10 = 55$. The opponents collect $T + 5$ runs in the losses (one extra run per loss) and $\tfrac{55-T}{2}$ runs in the doubling games, so the opponent total is $(T+5) + \tfrac{55-T}{2} = \tfrac{T+65}{2}$. This says something the step-by-step route does not: the answer depends only on $T$, not on which particular games were which. It remains to pin $T$, and one count does it — the doubling scores are five different even numbers taken from $1$ to $10$, and there are only five such numbers, so they total $30$ and $T = 55 - 30 = 25$. Then $\tfrac{25+65}{2} = 45$, matching the first route. The formula also explains why the total comes out whole: $\tfrac{T+65}{2}$ is a whole number only when $T$ is odd, and the forced split makes $T = 25$.
CCSS standards used (min grade 4)
4.OA.A.1Interpret a multiplication equation as a comparison (Reading "scored twice as many runs as their opponent" as $s = 2o$ and inverting it to $o = \tfrac{s}{2}$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that a doubling game forces the team's score to be a multiple of $2$, so only $2,4,6,8,10$ qualify.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (The counting argument that five doubling games and five available even scores force the split, plus the check that the resulting arrangement actually exists.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the opponent runs group by group: $2+4+6+8+10 = 30$, $1+2+3+4+5 = 15$, and $30+15 = 45$.)
⭐ Half of an odd number of runs is not a whole number, so the five odd scores had to be the one-run losses — and with only five even scores left over, there is exactly one way the other five games can go.
⭐ Half of an odd number of runs is not a whole number, so the five odd scores had to be the one-run losses — and with only five even scores left over, there is exactly one way the other five games can go.