AMC 10 · 2013 · #2

Grade 4 logiccounting
0
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Problem
Half the games were lost by one and in the rest the score was double the opponent's. Find the opponents' total.

Pick an answer.

(A)
35
(B)
40
(C)
45
(D)
50
(E)
55
How to solve
Strategy Eliminate Possibilities

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.

1STEP 1

Two game types, two opponent formulas

Each game type gives its own formula.

one-run loss: o = s + 1 doubling game: s = 2o → o = s/2
2STEP 2

A doubling game needs an even score

A doubling game needs an even score.

o = s/2 is a whole number ⇔ s is even ⇔ s ∈ {2,4,6,8,10}
3STEP 3

Five evens fill five slots exactly

The five evens fill the five slots exactly.

{doubling scores} ⊆ {2,4,6,8,10}, |{doubling scores}| = 5 = |{2,4,6,8,10}| → {doubling scores} = {2,4,6,8,10}
4STEP 4

Add the two group totals

Adding both groups gives 45, choice (C).

losses: 2+4+6+8+10 = 30 doubling: 1+2+3+4+5 = 15 30 + 15 = 45 → (C)
Answer
45
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 (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.
💡Key takeaway

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.

  • Two game types, two opponent formulas
  • A doubling game needs an even score
  • Five evens fill five slots exactly
  • Add the two group totals