2012 AMC 10B Problems/Problem 7

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 $lcm<cmath>3,4</cmath>=12$ , 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 $\textbf{48(D)}$

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2012 AMC 10B Problems/Problem 7

Contents

Problem 7

Solution 1

Solution 2

See Also