1989 AMC 8 Problem 2

Attempt Problem 2 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).

2.

What is the value of
210+4100+61000?\frac{2}{10} + \frac{4}{100} + \frac{6}{1000}?
0.0120.012
0.02460.0246
0.120.12
0.2460.246
246246
Answer: D
Concepts:decimalplace value
Small hint:
Write each fraction as a decimal using its place value
Big hint:
210=0.2,\frac{2}{10} = 0.2, 4100=0.04,\frac{4}{100} = 0.04, and 61000=0.006\frac{6}{1000} = 0.006
Solution:
Each fraction is a decimal in a different place: 210=0.2,\frac{2}{10} = 0.2, 4100=0.04,\frac{4}{100} = 0.04, and 61000=0.006.\frac{6}{1000} = 0.006.
Adding these gives 0.2+0.04+0.006=0.246.0.2 + 0.04 + 0.006 = 0.246.
Thus, the correct answer is D .
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