Problem 21 · 2022 AMC 8
Hard
Fractions, Decimals & Percents
percent-multipliercasework
Steph scored 15 baskets out of 20 attempts in the first half of a game, and 10 baskets out of 10 attempts in the second half. Candace took 12 attempts in the first half and 18 attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?
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Answer: C — 9 more baskets.
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Hint 1 of 2
Both players took the same number of total attempts (30). If two people with equal attempts finish at the same overall percentage, what must be equal about their makes? That collapses the “surprising” clue into a hard number.
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Hint 2 of 2
Candace also made 25 baskets. Now her per-half percentages are each strictly below Steph's, which caps her first-half and second-half makes — and only one split of 25 fits both caps.
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Approach: equal attempts + equal overall % forces equal total makes; then squeeze the split
- Insight: turn the “surprising” tie into arithmetic. Both shot 30 total (Steph 20+10, Candace 12+18). Equal attempts and equal overall percentage means equal makes — Steph made 15 + 10 = 25, so Candace made 25 too.
- Let Candace's makes be f (of 12) and s (of 18), with f + s = 25. Beating-by-Steph in each half caps her: f/12 < 15/20 = ¾ forces f ≤ 8, and s/18 < 1 forces s ≤ 17.
- Those caps add to exactly 8 + 17 = 25, so the only split is f = 8, s = 17 — any less in one half can't be made up in the other.
- s − f = 17 − 8 = 9.
- Why the caps pin it down: when two upper bounds sum to exactly the required total, each variable is pinned to its max — no slack to trade. (This is also the resolution of the classic “Simpson's paradox” setup: losing both halves yet tying overall.)
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