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== Problem 7 ==
 
== 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==

Revision as of 18:19, 5 August 2020
































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 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 $\textbf{48(D)}$

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. AMC logo.png

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