2010 AMC 10A Problem 1

Attempt Problem 1 of the 2010 AMC 10A 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 2010 AMC 10A solutions, or check the answer key.

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

1.

Mary’s top book shelf holds five books with the following widths, in centimeters: 6,6, 12,\dfrac{1}{2}, 1,1, 2.5,2.5, and 10.10.
What is the average book width, in centimeters?
11
22
33
44
55
Answer: D
Concepts:meandecimal
Small hint:
Add the widths before dividing by the number of books
Big hint:
Compute (6+12+1+2.5+10)÷5\left(6+\dfrac12+1+2.5+10\right)\div5
Solution:
Adding the five widths gives 6+12+1+2.5+10=20.6+\dfrac12+1+2.5+10=20. The average is 20÷5=4.20\div5=4.
Thus, D is the correct answer.
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