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2010 AMC 10A Problem 4

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2010 AMC 10A Problem 4AMC 10 Step by Step
Name:                             Date: September 13, 2026
Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

A book that is to be recorded onto compact discs takes 412412 minutes to read aloud. Each disc can hold up to 5656 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?

  1. A)

    50.250.2

  2. B)

    51.551.5

  3. C)

    52.452.4

  4. D)

    53.853.8

  5. E)

    55.255.2

Answer

B

Key insight

412/56 is a bit more than 7, so 8 discs are needed, and then each disc holds 412/8 = 51.5 minutes.

Solution

First find how many discs are needed. Since 7⋅56=392<4127 \cdot 56 = 392 < 412 and 8⋅56=448≥4128 \cdot 56 = 448 \ge 412, seven discs are not enough and eight are. So 88 discs are used.

Spreading 412412 minutes evenly over 88 discs puts
4128=51.5\frac{412}{8} = 51.5
minutes on each disc, which is comfortably under the 5656-minute limit.

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

Why this works

"Smallest number of containers" is a ceiling computation: divide and round up. Only after the count is fixed does the "same length on each" condition turn into an ordinary division. Two separate questions are hidden in one sentence; answer them in order.

The trap

Rounding 412/56 down to 7 discs, or answering 56 minutes because that is the capacity.

Common mistakes

  • Rounding 412/56 down to 7 discs, or answering 56 minutes because that is the capacity.
  • Computing 412/7≈58.9412/7 \approx 58.9, which exceeds the disc capacity and should have signalled the error.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed