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

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.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Page

Toolbox

Search

2015 AMC 8 Problems/Problem 20

Solution

See Also