Difference between revisions of "2012 AMC 10B Problems/Problem 7"
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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math> | <math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math> | ||
| + | ==Solution 1== | ||
| + | Let <math>x</math> be the number of acorns that both animals had. | ||
| − | = | + | So by the info in the problem: |
| − | x | + | |
| + | <math>\frac{x}{3}=\left( \frac{x}{4} \right)+4</math> | ||
| + | |||
| + | Subtracting <math> \frac{x}{4}</math> from both sides leaves | ||
| − | |||
<math>\frac{x}{12}=4</math> | <math>\frac{x}{12}=4</math> | ||
| + | |||
<math>\boxed{x=48}</math> | <math>\boxed{x=48}</math> | ||
| − | + | This is answer choice <math>\textbf{(D)}</math> | |
| + | |||
| + | ==Solution 2== | ||
| + | Instead of an Algebraic Solution, we can just find a residue in the holes through common multiples of <math>3</math> and <math>4</math>, so <math>lcm[3,4]=12</math>, the next largest is <math>12\cdot2=24</math>, the next is <math>36</math>, and so on, with all of them being multiples of <math>12</math>, now we can see that per every common multiple, we can see a pattern such as | ||
| + | |||
| + | <math>12=4\cdot3=3\cdot4</math> so <math>4-3=1</math> hole less. | ||
| + | |||
| + | <math>24=4\cdot6=3\cdot8</math> so <math>8-6=2</math> holes less. | ||
| + | |||
| + | <math>36=4\cdot9=3\cdot12</math> so <math>12-9=3</math> holes less. | ||
| + | |||
| + | <math>48=4\cdot12=3\cdot16</math> so <math>16-12=4</math> holes less. | ||
| + | |||
| + | So we see that <math>48</math> is the number we need which is <math>\textbf{48(D)}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/hRlDVKgAv9U | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| − | + | {{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 15:43, 5 October 2022
Problem 7
For a science project, Sammy observed a chipmunk and a 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?
Solution 1
Let 
be the number of acorns that both animals had.
So by the info in the problem:
Subtracting 
from both sides leaves
This is answer choice 
Solution 2
Instead of an Algebraic Solution, we can just find a residue in the holes through common multiples of 
and 
, so ![$lcm[3,4]=12$](http://latex.artofproblemsolving.com/1/a/d/1ad88f751508c871c757a7654863e5c575b85f58.png)
, the next largest is 
, the next is 
, and so on, with all of them being multiples of 
, now we can see that per every common multiple, we can see a pattern such as
so 
hole less.
so 
holes less.
so 
holes less.
so 
holes less.
So we see that 
is the number we need which is 
Video Solution
~savannahsolver
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
