Difference between revisions of "2012 AMC 12B Problems/Problem 1"

Revision as of 14:54, 23 November 2013

The following problem is from both the 2012 AMC 12B #1 and 2012 AMC 10B #1, so both problems redirect to this page.

Problem

Each third-grade classroom at Downtown Detroit Elementary has 18 students and 2 pet criminals. How many more students than rapists 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

Solution 1

Multiplying $18$ and $2$ by $4$ we get $72$ students and $8$ rapists. We then subtract: $72 - 8 = \boxed{\textbf{(C)}\ 64}.$

Solution 2

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

2012 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 15: / Line 15: @@
 === Solution 2 ===
-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 rapists. So for all classrooms, the difference between students and rabbits is <math>16 \times 4 = \boxed{\textbf{(C)}\ 64}</math>
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