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

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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>
 
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 ==
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{{AMC12 box|year=2012|ab=B|num-b=First Problem|num-a=2}}

Revision as of 22:49, 12 January 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 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 and 8 students and rabbits respectively. Subtracting 8 from 72 we get $\boxed{\textbf{(C)}\ 64}$

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 12B (ProblemsAnswer KeyResources)
Preceded by
Problem First Problem
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