1999 AMC 12 Problem 3

Attempt Problem 3 of the 1999 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 1999 AMC 12 solutions, or check the answer key.

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

3.

The number halfway between 18\tfrac18 and 110\tfrac{1}{10} is
180\dfrac{1}{80}
140\dfrac{1}{40}
118\dfrac{1}{18}
19\dfrac{1}{9}
980\dfrac{9}{80}
Answer: E
Concepts:fractionmean
Small hint:
The number halfway between two values is their average
Big hint:
Compute 12(18+110)\dfrac12\left(\dfrac18 + \dfrac{1}{10}\right)
Solution:
The halfway point is the average 12(18+110)=12⋅1880=980. \dfrac12\left(\dfrac18 + \dfrac{1}{10}\right) = \dfrac12 \cdot \dfrac{18}{80} = \dfrac{9}{80}.
Thus, the correct answer is E.
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