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

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 $18$ 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 $16 \times 4 = \boxed{\textbf{(C)}\ 64}$

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.

@@ Line 19: / Line 19: @@
 == See Also ==
-{{AMC10 box|year=2012|ab=B|before=|num-a=2}}
+{{AMC10 box|year=2012|ab=B|before=First Question|num-a=2}}
-{{AMC12 box|year=2012|ab=B|before=|num-a=2}}
+{{AMC12 box|year=2012|ab=B|before=First Question|num-a=2}}
 {{MAA Notice}}

Page

Toolbox

Search