# Difference between revisions of "2015 AMC 8 Problems/Problem 20"

Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost$1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy? $\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

## Solution 1

So let there be $x$ pairs of $1$ dollar socks, $y$ pairs of $3$ dollar socks, $z$ pairs of $4$ dollar socks.

We have $x+y+z=12$, $x+3y+4z=24$, and $x,y,z \ge 1$.

Now we subtract to find $2y+3z=12$, and $y,z \ge 1$. It follows that $y$ is a multiple of $3$ and $2y$ is a multiple of $6$, so since $0<2y<12$, we must have $2y=6$.

Therefore, $y=3$, and it follows that $z=2$. Now $x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}$, as desired.

## Solution 2

Since the total cost of the socks was $24$ and Ralph bought $12$ pairs, the average cost of each pair of socks is $\frac{24}{12} = 2$.

There are two ways to make packages of socks that average to $2$. You can have: $\bullet$ Two $1$ pairs and one $4$ pair (package adds up to $6$) $\bullet$ One $1$ pair and one $3$ pair (package adds up to $4$)

So now we need to solve $$6a+4b=24,$$ where $a$ is the number of $6$ packages and $b$ is the number of $4$ packages. We see our only solution (that has at least one of each pair of sock) is $a=2, b=3$, which yields the answer of $2\times2+3\times1 = \boxed{\textbf{(D)}~7}$.

## See Also

 2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 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 AJHSME/AMC 8 Problems and Solutions

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