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2012 AMC 10B Problem 1

2012 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: October 28, 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?

A) $48$
B) $56$
C) $64$
D) $72$
E) $80$

Answer

Key insight

Each classroom has 16 more students than rabbits. There are 4 classrooms, so there are 4 times 16, which is 64, more in all.

Solution

Look at one classroom first. It has $18$ students and $2$ rabbits, so it has $18 - 2 = 16$ more students than rabbits.
The four classrooms are all the same, so each one adds another $16$ to the gap.
Four classrooms give $4 \times 16 = 64$ .

Check it another way: there are $4 \times 18 = 72$ students and $4 \times 2 = 8$ rabbits, and $72 - 8 = 64$ . Both ways agree.

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

Why this works

"How many more" asks for a difference. When every room has the same difference, the total difference is just that one-room difference times the number of rooms. Working out one room first keeps the numbers small.

The trap

Multiplying the students by 4 to get 72 and then forgetting to take away the rabbits, which gives the wrong choice (D).

Concise solution

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

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

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)

Techniques

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