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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>
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==Solution 1==
 
==Solution 1==
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==Solution 2==
 
==Solution 2==
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Instead of an Algebraic Solution, we can just find a residue in the common multiples of <math>3</math> and <math>4</math>, so <math>lcm<cmath>3,4</cmath>=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
+
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  
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<math>12=4\cdot3=3\cdot4</math> so <math>4-3=1</math> hole less.
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<math>24=4\cdot6=3\cdot8</math> so <math>8-6=2</math> holes less.
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<math>12=4\cdot3=3\cdot4</math> so <math>4-3=1</math> hole less.  
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<math>36=4\cdot9=3\cdot12</math> so <math>12-9=3</math> holes less.
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<math>48=4\cdot12=3\cdot16</math> so <math>16-12=4</math> holes less.
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<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>
 
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==
 
==See Also==
  
 
{{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}}
 
{{AMC10 box|year=2012|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{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?

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

Solution 1

Let $x$ be the number of acorns that both animals had.

So by the info in the problem:

$\frac{x}{3}=\left( \frac{x}{4} \right)+4$

Subtracting $\frac{x}{4}$ from both sides leaves

$\frac{x}{12}=4$

$\boxed{x=48}$

This is answer choice $\textbf{(D)}$

Solution 2

Instead of an Algebraic Solution, we can just find a residue in the holes through common multiples of $3$ and $4$, so $lcm[3,4]=12$, the next largest is $12\cdot2=24$, the next is $36$, and so on, with all of them being multiples of $12$, now we can see that per every common multiple, we can see a pattern such as

$12=4\cdot3=3\cdot4$ so $4-3=1$ hole less.

$24=4\cdot6=3\cdot8$ so $8-6=2$ holes less.

$36=4\cdot9=3\cdot12$ so $12-9=3$ holes less.

$48=4\cdot12=3\cdot16$ so $16-12=4$ holes less.

So we see that $48$ is the number we need which is $\textbf{48(D)}$

Video Solution

https://youtu.be/hRlDVKgAv9U

~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

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