1995 AHSME Problems/Problem 16

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Problem

Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:

i. The actual attendance in Atlanta is within $10 \%$ of Anita's estimate.

ii. Bob's estimate is within $10 \%$ of the actual attendance in Boston.

To the nearest 1,000, the largest possible difference between the numbers attending the two games is

$\mathrm{(A) \ 10000 } \qquad \mathrm{(B) \ 11000 } \qquad \mathrm{(C) \ 20000 } \qquad \mathrm{(D) \ 21000 } \qquad \mathrm{(E) \ 22000 }$

Solution

Since the number of people at the game in Boston is certainly more than the number of fans in Atlanta, we need to compute the maximum of Bob's game minus the minimum of Anita's game. Note however that there is a slight different between conditions (i) and (ii); the attendence is within $\pm 10 \%$ from Anita's estimate of $50,000$ , but Bob's estimate is within $\pm 10 \%$ of the actual attendance, or $0.9x \le 60000 \le 1.1x \Longrightarrow x \le \frac{60000}{0.9}$ .

The answer is $\frac{60000}{0.9} - 50000 \times 0.9 \approx 22000 \Longrightarrow \mathrm{(E)}$ .

1995 AHSME (Problems)
Preceded by Problem 15	Followed by Problem 17
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1995 AHSME Problems/Problem 16

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