Printable

2010 AMC 10A Problem 1

2010 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: October 28, 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

Average means total divided by count. Add in easy groups: 6 + 10 = 16 and 1/2 + 2.5 + 1 = 4, so 20 / 5 = 4.

Solution

The average width is the total width divided by the number of books.

There are $5$ books, so count them first: $6$ , $\tfrac12$ , $1$ , $2.5$ , $10$ .
Add them in groups that give whole numbers. The two big ones: $6 + 10 = 16$ . The three small ones: $\tfrac12 + 2.5 = 3$ , and $3 + 1 = 4$ .
The total is $16 + 4 = 20$ centimeters.
Share the total over the $5$ books: $20 \div 5 = 4$ .

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

Why this works

An average is always "add them all up, then divide by how many there are." Adding in friendly groups, such as a half plus two and a half, keeps the arithmetic short and lowers the chance of a slip, even on an easy problem.

The trap

Dividing by 4 instead of 5 (forgetting a book), or making a slip when adding 1/2 and 2.5.

Concise 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 $\boxed{\textbf{(D)}\ 4}$ .

Review in the AoPS books

Prealgebra, Chapter 13: Data and Statistics

Techniques

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