1989 AMC 8 Problem 4

Attempt Problem 4 of the 1989 AMC 8 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 1989 AMC 8 solutions, or check the answer key.

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

4.

Estimate to determine which of the following is closest to
4010.205.\frac{401}{0.205}.
0.20.2
22
2020
200200
20002000
Answer: E
Concepts:estimation
Small hint:
Round 401401 and 0.2050.205 to simple numbers
Big hint:
4010.205≈4000.2\frac{401}{0.205} \approx \frac{400}{0.2}
Solution:
Round the numerator and denominator: 4010.205≈4000.2.\frac{401}{0.205} \approx \frac{400}{0.2}.
Since 4000.2=2000,\frac{400}{0.2} = 2000, the value is closest to 2000.2000.
Thus, the correct answer is E .
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