2012 AMC 10B Problem 1

P1-10~1 minPrint

Linear Equations & Word Problems Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

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.

Concise solution

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

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)