Difference between revisions of "2012 AMC 12B Problems/Problem 3"

(Problem)
 
(2 intermediate revisions by one other user not shown)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
If x is the number of holes that the chipmunk dug then, 3x=4*(x-4). so 3x=4x-16. x=16, and to find number of acorns hid by chipmunk, multiply by three, to get 48; D.
+
If <math>x</math> is the number of holes that the chipmunk dug, then <math>3x=4(x-4)</math>, so <math>3x=4x-16</math>, and <math>x=16</math>. The number of acorns hidden by the chipmunk is equal to <math>3x = \boxed{\textbf{(D)}\ 48}</math>
  
 
== See Also ==
 
== See Also ==
Line 13: Line 13:
  
 
{{AMC12 box|year=2012|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2012|ab=B|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Latest revision as of 15:57, 3 July 2013

Problem

For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?

$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$

Solution

If $x$ is the number of holes that the chipmunk dug, then $3x=4(x-4)$, so $3x=4x-16$, and $x=16$. The number of acorns hidden by the chipmunk is equal to $3x = \boxed{\textbf{(D)}\ 48}$

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

Invalid username
Login to AoPS