Difference between revisions of "2012 AMC 12B Problems/Problem 1"
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| + | {{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #1]] and [[2012 AMC 10B Problems|2012 AMC 10B #1]]}} | ||
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== Problem == | == Problem == | ||
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== Solution == | == Solution == | ||
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| + | === Solution 1 === | ||
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Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get | Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get | ||
<math>\boxed{\textbf{(C)}\ 64}</math> | <math>\boxed{\textbf{(C)}\ 64}</math> | ||
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| + | === Solution 2 === | ||
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| + | 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> | ||
Revision as of 15:39, 23 August 2012
- The following problem is from both the 2012 AMC 12B #1 and 2012 AMC 10B #1, so both problems redirect to this page.
Contents
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?
Solution
Solution 1
Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get
Solution 2
In each class, there are 
more students than rabbits. So for all classrooms, the difference between students and rabbits is 
