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

m
Line 11: Line 11:
 
=== Solution 1 ===
 
=== Solution 1 ===
  
Multiplying 18 and 2 by 4 we get 72 and 8 students and rabbits respectively. Subtracting 8 from 72 we get
+
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 ===

Revision as of 13:02, 5 July 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 $72$ 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png