Difference between revisions of "2012 AMC 12B Problems/Problem 1"
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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> | ||
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=== Solution 1 === | === Solution 1 === | ||
| − | Multiplying 18 and 2 by 4 we get 72 and 8 | + | Multiplying <math>18</math> and <math>2</math> by <math>4</math> we get <math>72</math> students and <math>8</math> rabbits. We then subtract: <math>72 - 8 = \boxed{\textbf{(C)}\ 64}.</math> |
| − | <math>\boxed{\textbf{(C)}\ 64}</math> | ||
=== Solution 2 === | === Solution 2 === | ||
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== See Also == | == See Also == | ||
| − | {{AMC12 box|year=2012|ab=B|before=First | + | {{AMC10 box|year=2012|ab=B|before=First Question|num-a=2}} |
| + | {{AMC12 box|year=2012|ab=B|before=First Question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:05, 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 
students and 
pet rabbits. How many more students than rabbits are there in all 
of the third-grade classrooms?
Solution
Solution 1
Multiplying and by we get 
students and 
rabbits. We then subtract: 
Solution 2
In each class, there are 
more students than rabbits. So for all classrooms, the difference between students and rabbits is 
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
