2012 AMC 12B Problems/Problem 1

Revision as of 20:05, 8 February 2014 by Flamedragon (talk | contribs) (See Also)
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)
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Followed by
Problem 2
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All AMC 10 Problems and Solutions
2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
'
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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