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2009 AMC 10A Problem 1

2009 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

One can holds $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ( $128$ ounces) of soda?

A) $7$
B) $8$
C) $9$
D) $10$
E) $11$

Answer

Key insight

Ten cans hold only 120 ounces, which is 8 short, so one more can is needed even though it will not be full.

Solution

We want the fewest cans that together hold at least $128$ ounces. Count up by whole cans.

Ten cans hold $10 \times 12 = 120$ ounces. That is $128 - 120 = 8$ ounces short, so ten cans are not enough.
Eleven cans hold $11 \times 12 = 132$ ounces, which is more than $128$ . So eleven cans are enough.
The eleventh can is only partly used (it holds the last $8$ ounces), but we still need it.

The answer is $\boxed{\textbf{(E)}\ 11}$ .

Why this works

When something comes in fixed-size containers and you need at least a certain amount, divide and then round up. Any leftover amount, even a small one, needs one more container. Rounding down would leave you short.

The trap

Working out 128 divided by 12 as about 10.7 and rounding down to 10 cans, which gives only 120 ounces.

Concise solution

We need the smallest whole number $n$ with $12n \ge 128$ .

Since $12 \cdot 10 = 120 < 128$ , ten cans are not enough. Since $12 \cdot 11 = 132 \ge 128$ , eleven cans suffice.

Equivalently, $128 = 12 \cdot 10 + 8$ : ten full cans plus $8$ more ounces, and those $8$ ounces require an eleventh can.

The answer is $\boxed{\textbf{(E)}\ 11}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)

Techniques

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