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Difference between revisions of "2012 AMC 10B Problems/Problem 1"

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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 13:14, 4 July 2013

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 $16 \times 4 = \boxed{\textbf{(C)}\ 64}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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