1995 AMC 12 Problem 16

Attempt Problem 16 of the 1995 AMC 12 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 1995 AMC 12 solutions, or check the answer key.

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

16.

Anita attends a baseball game in Atlanta and estimates that there are 50,00050{,}000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,00060{,}000 fans in attendance. A league official who knows the actual numbers attending the two games notes that:
i. The actual attendance in Atlanta is within 10%10\% of Anita’s estimate.
ii. Bob’s estimate is within 10%10\% of the actual attendance in Boston.
To the nearest 1,000,1{,}000, the largest possible difference between the numbers attending the two games is
10,00010{,}000
11,00011{,}000
20,00020{,}000
21,00021{,}000
22,00022{,}000
Answer: E
Concepts:percent errorinequalities
Small hint:
In the two statements, the 10%10\% is taken of different quantities
Big hint:
Find both attendance intervals, then maximize the larger endpoint minus the smaller endpoint
Solution:
Atlanta’s actual attendance AA satisfies 45,000≤A≤55,000.45{,}000\le A\le55{,}000. For Boston’s actual attendance B,B, the condition is ∣60,000−B∣≤0.1B,|60{,}000-B|\le0.1B, so 60,0001.1≤B≤60,0000.9.\frac{60{,}000}{1.1}\le B\le\frac{60{,}000}{0.9}. The largest difference is therefore 60,0000.9−45,000=21,666.6‾, \frac{60{,}000}{0.9}-45{,}000 =21{,}666.\overline6, which rounds to 22,000.22{,}000. Thus the correct answer is E.
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