2010 AMC 12A Problem 5

Attempt Problem 5 of the 2010 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

Halfway through a 100100-shot archery tournament, Chelsea leads by 5050 points. For each shot a bullseye scores 1010 points, with other possible scores being 8,8, 4,4, 2,2, and 00 points. Chelsea always scores at least 44 points on each shot. If Chelsea’s next nn shots are bullseyes she will be guaranteed victory. What is the minimum value for n?n?
3838
4040
4242
4444
4646
Answer: C
Concepts:inequalitybounding to limit cases
Small hint:
Chelsea’s opponent can score at most 1010 points on each of the remaining 5050 shots
Big hint:
With nn bullseyes and at least 44 points on the other 50−n50-n shots, require 10n+4(50−n)>500−5010n+4(50-n)\gt 500-50
Solution:
The opponent can score at most 50⋅10=50050\cdot10=500 on the last 5050 shots. Since Chelsea leads by 50,50, she must score more than 500−50=450500-50=450 points on her remaining shots to guarantee victory.
Her nn bullseyes give 10n10n points, and her other 50−n50-n shots give at least 4(50−n)4(50-n) points, so 10n+4(50−n)>450.10n+4(50-n)\gt450. This simplifies to 6n>250,6n\gt250, i.e. n>4123.n\gt41\tfrac23.
Therefore Chelsea needs at least 4242 bullseyes.
Thus, C is the correct answer.
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