2009 AMC 10A Problem 1

P1-10~1 minPrint

Linear Equations & Word Problems Number Properties Set up the equation/formula and compute; no special trick needed

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

Show solution without answering

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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)