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2012 AMC 10B Problem 1

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2012 AMC 10B 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

Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?

  1. A)

    4848

  2. B)

    5656

  3. C)

    6464

  4. D)

    7272

  5. E)

    8080

Answer

C

Key insight

Each classroom has 18 - 2 = 16 more students than rabbits, so four classrooms have 4 * 16 = 64 more.

Solution

In a single classroom there are 18−2=1618 - 2 = 16 more students than rabbits. The four classrooms are identical, so the total excess is
4⋅16=64.4 \cdot 16 = 64.

The answer is (C) 64\boxed{\textbf{(C)}\ 64}.

Why this works

"How many more" is a difference, and a difference that is the same in every room simply multiplies by the number of rooms. Equivalently, 4⋅18−4⋅2=72−8=644 \cdot 18 - 4 \cdot 2 = 72 - 8 = 64; the distributive property says both routes agree.

The trap

Multiplying only the students by 4 (giving 72) and subtracting just 2 rabbits, or forgetting the rabbits altogether.

Common mistakes

  • Multiplying only the students by 4 (giving 72) and subtracting just 2 rabbits, or forgetting the rabbits altogether.
  • Adding instead of subtracting, giving 4⋅20=804 \cdot 20 = 80, choice (E).

Techniques

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