2010 AMC 10A Problem 4

P1-10~1 minPrint

Linear Equations & Word Problems Number Properties Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

A book that is to be recorded onto compact discs takes $412$ minutes to read aloud. Each disc can hold up to $56$ minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?

Show solution without answering

Key insight

First decide how many discs are needed: 7 discs hold only 392 minutes, so 8 are needed. Then share 412 minutes equally: 412 / 8 = 51.5.

Solution

The question hides two steps: first find how many discs are used, then share the reading evenly.

1. How many discs? Each disc holds at most $56$ minutes. Try whole numbers of discs:
- $7$ discs hold at most $7 \times 56 = 392$ minutes. That is less than $412$ , so $7$ is not enough.
- $8$ discs hold up to $8 \times 56 = 448$ minutes. That is enough.

So the smallest number of discs is $8$ .
2. How much on each? The $412$ minutes are split equally over $8$ discs:
$412 \div 8 = 51.5 \text{ minutes}.$
3. Check: $51.5$ is under the $56$ -minute limit, so each disc can hold it.

The answer is $\boxed{\textbf{(B)}\ 51.5}$ .

Why this works

"Fewest containers" means divide and round up: $412 \div 56$ is about $7.4$ , and you cannot use part of a disc, so you need $8$ . Once the number of discs is fixed, "the same amount on each" is a plain division. Many word problems pack two small questions into one sentence; answer them one at a time.

Concise solution

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

First find how many discs are needed. Since $7 \cdot 56 = 392 < 412$ and $8 \cdot 56 = 448 \ge 412$ , seven discs are not enough and eight are. So $8$ discs are used.

Spreading $412$ minutes evenly over $8$ discs puts
$\frac{412}{8} = 51.5$
minutes on each disc, which is comfortably under the $56$ -minute limit.

The answer is $\boxed{\textbf{(B)}\ 51.5}$ .

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)