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

(Solution 2)
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</math>
 
</math>
  
==Solution==
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==Solution 1==
 
So let there be <math>x</math> pairs of <math>1</math> dollar socks, <math>y</math> pairs of <math>3</math> dollar socks, <math>z</math> pairs of <math>4</math> dollar socks.
 
So let there be <math>x</math> pairs of <math>1</math> dollar socks, <math>y</math> pairs of <math>3</math> dollar socks, <math>z</math> pairs of <math>4</math> dollar socks.
  
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Therefore, <math>y=3</math>, and it follows that <math>z=2</math>. Now <math>x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}</math>, as desired.
 
Therefore, <math>y=3</math>, and it follows that <math>z=2</math>. Now <math>x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}</math>, as desired.
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==Solution 2==
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Since the total cost of the socks was <math>\$24</math> and Ralph bought <math>12</math> pairs, the average cost of each pair of socks is <math>\frac{\$24}{12} = \$2</math>.
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There are two ways to make packages of socks that average to <math>\$2</math>. You can have:
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<math>\bullet</math> Two <math>\$1</math> pairs and one <math>\$4</math> pair (package adds up to <math>\$6</math>)
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<math>\bullet</math> One <math>\$1</math> pair and one <math>\$3</math> pair (package adds up to <math>\$4</math>)
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So now we need to solve
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<cmath>6a+4b=24,</cmath>
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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 Also==
  
 
{{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 01:59, 27 November 2015

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 (ProblemsAnswer KeyResources)
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

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