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 $$1$ socks, $y$ pairs of $$3$ socks, $z$ pairs of $$4$ 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 $2\times2+3\times1 = \boxed{\textbf{(D)}~7}$ .

Solution 3

Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$ . This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\cdot1=9$ dollars wasted, which leaves $7$ more dollars. If we replace one pair with a $3$ dollar pair, then we would waste an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would waste an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$ s and $3$ s is $2+2+3$ . If we change $3$ pairs, we would have $6$ pairs left. Adding the one pair from previously, we have $\boxed{(\text{D})~7}$ pairs.

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2015 AMC 8 Problems/Problem 20

Contents

Solution 1

Solution 2

Solution 3

See Also