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2009 AMC 10A Problem 1

2009 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: September 13, 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

Divide 128 by 12 and round up, because a partially used can still counts as a whole can.

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 $\boxed{\textbf{(E)}\ 11}$ .

Why this works

"Minimum number of containers" is a ceiling question: divide, and if there is any remainder at all, add one more container. Whenever a quantity must be at least something and comes in fixed-size units, round the quotient up, never down.

The trap

Rounding 128/12 = 10.67 down to 10 cans, which supplies only 120 ounces and falls 8 ounces short.

Common mistakes

Rounding 128/12 = 10.67 down to 10 cans, which supplies only 120 ounces and falls 8 ounces short.
Misremembering the size of a gallon; the problem gives $128$ ounces, so use it directly.

Techniques

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