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

m (Solution 1)
m (Solution 2)
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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>)
 
<math>\bullet</math> One <math>$1</math> pair and one <math>$3</math> pair (package adds up to <math>$4</math>)
  
So now we need to solve
+
Now, we need to solve
 
<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>.

Revision as of 05:34, 4 May 2023

Problem

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?

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solutions

Solution 1

So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ 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 $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$. Since sum of 2 multiples of 3 = multiple of 3, so 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$)

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.

Solution 4

Let the amount of $1$ dollar socks be $a$, $3$ dollar socks be $b$, and $4$ dollar socks be $c$. We then know that $a+b+c=12$ and $a+3b+4c=24$. We can make $a+b+c=12$ into $a=12-b-c$ and then plug that into the other equation, producing $12-b-c+3b+4c=24$ which simplifies to $2b+3c=12$. It's not hard to see $b=3$ and $c=2$. Now that we know $b$ and $c$, we know that $a=7$, meaning the number of $1$ dollar socks Ralph bought is $\boxed{\textbf{(D)} 7}$

Video Solution

https://youtu.be/hvnVuLbveJs

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=2187

~pi_is_3.14

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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