2010 AMC 10A Problem 4

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 $\boxed{\textbf{(B)}\ 51.5}$.

"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.

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 $\boxed{\textbf{(B)}\ 51.5}$.