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

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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>.
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==Solution 3==
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Since there are 12 pairs socks, and Ralph bought at least one pair of each, there are <math>12-3=9</math> pairs of socks left. Also, the sum of the three pairs of socks is <math>1+3+4=8</math>. This means that there are <math>24-8=16</math>dollars left. If there are only <math>1</math> dollar socks left, then we would have <math>9\cdot1=9</math> dollars wasted, which leaves <math>7</math> more dollars. If we replace one pair with a <math>3</math> dollar pair, then we would waste an additional <math>2</math> dollars. If we replace one pair with a <math>4</math> dollar pair, then we would waste an additional <math>3</math> dollars. The only way <math>7</math> can be represented as a sum of <math>2</math>s and <math>3</math>s is <math>2+2+3</math>. If we change <math>3</math> pairs, we would have <math>6</math> pairs left. Adding the one pair from previously, we have <math>\boxed{(\text{D})~7}</math> pairs.
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==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 16:32, 28 October 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 $2\times2+3\times1 = \boxed{\textbf{(D)}~7}$.

Solution 3

Since there are 12 pairs 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.

See Also

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

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