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2010 AMC 10A Problem 1

2010 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 1 of 5 · about 1 min · position P1-10

Mary’s top book shelf holds five books with the following widths, in centimeters: $6$ , $\dfrac{1}{2}$ , $1$ , $2.5$ , and $10$ . What is the average book width, in centimeters?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

Add the five widths, which pair up nicely (6 + 10 = 16, 2.5 + 0.5 + 1 = 4), then divide by 5 to get 20/5 = 4.

Solution

The average is the total width divided by the number of books. Group the numbers so the fractions cancel:
$6 + 10 = 16, \qquad \tfrac12 + 2.5 + 1 = 4,$
so the total is $16 + 4 = 20$ centimeters. Dividing by the $5$ books gives
$\frac{20}{5} = 4.$

The answer is $\boxed{\textbf{(D)}\ 4}$ .

Why this works

The mean of a list is nothing more than its sum divided by its length. Pairing values that combine to whole numbers keeps the arithmetic clean and error-free, a habit worth keeping even on the easiest problems.

The trap

Adding the fractions carelessly, or dividing by 4 instead of by 5 books.

Common mistakes

Adding the fractions carelessly, or dividing by 4 instead of by 5 books.
Mixing up $\tfrac12$ and $2.5$ when converting to decimals, which shifts the total away from $20$ .

Techniques

Set up the equation/formula and compute; no special trick needed