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2012 AMC 10B Problems/Problem 1: Difference between revisions

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== Solution==
== Solution==
In each class, there are <math>18-2=16</math> more students than rabbits. So for all classrooms, the difference between students and rabbits is <math>16 \times 4 = \boxed{\textbf{(C)}\ 64}</math>
In each class, there are <math>18-2=16</math> more students than rabbits. So for all classrooms, the difference between students and rabbits is <math>16 \times 4 = \boxed{\textbf{(C)}\ 64}</math>
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{{MAA Notice}}

Revision as of 18:24, 15 February 2016

Problem

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

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$

Solution

In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \times 4 = \boxed{\textbf{(C)}\ 64}$

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