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2015 AMC 8 Problems/Problem 20: Difference between revisions

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<cmath>6a+4b=24,</cmath>
<cmath>6a+4b=24,</cmath>
where <math>a</math> is the number of <math>\$6</math> packages and <math>b</math> is the number of <math>\$4</math> packages. We see our only solution (that has at least one of each pair of sock) is <math>a=2, b=3</math>, which yields the answer of <math>2\times2+3\times1 = \boxed{\textbf{(D)}~7}</math>.
where <math>a</math> is the number of <math>\$6</math> packages and <math>b</math> is the number of <math>\$4</math> packages. We see our only solution (that has at least one of each pair of sock) is <math>a=2, b=3</math>, which yields the answer of <math>2\times2+3\times1 = \boxed{\textbf{(D)}~7}</math>.
==See Also==
==See Als==


{{AMC8 box|year=2015|num-b=19|num-a=21}}
{{AMC8 box|year=2015|num-b=19|num-a=21}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 15:00, 12 September 2020

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 $2\times2+3\times1 = \boxed{\textbf{(D)}~7}$.

See Als

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

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