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

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


<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 </math>
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 </math>

Revision as of 00:07, 20 January 2015

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 $16 \times 4 = \boxed{\textbf{(C)}\ 64}$ These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination