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

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== Problem ==
 
== Problem ==
 
   
 
   
Each third-grade classroom at Downtown Detroit Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?
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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>
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== See Also ==
 
== See Also ==
  
{{AMC12 box|year=2012|ab=B|before=|num-a=2}}
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{{AMC10 box|year=2012|ab=B|before=First Question|num-a=2}}
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{{AMC12 box|year=2012|ab=B|before=First Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:05, 8 February 2014

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 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

Solution 1

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

Solution 2

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

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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