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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| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2012|ab=B|num-b=1|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 12:32, 16 February 2016
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
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 Problem 1 |
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
