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

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==Problem==
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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?
 
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?
  
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</math>
 
</math>
  
==Solution 1==
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==Solutions==
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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===Solution 1===
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So, let there be <math>x</math> pairs of <math>\$1</math> socks, <math>y</math> pairs of <math>\$3</math> socks, and <math>z</math> pairs of <math>\$4</math> socks.
  
 
We have <math>x+y+z=12</math>, <math>x+3y+4z=24</math>, and <math>x,y,z \ge 1</math>.
 
We have <math>x+y+z=12</math>, <math>x+3y+4z=24</math>, and <math>x,y,z \ge 1</math>.
  
Now we subtract to find <math>2y+3z=12</math>, and <math>y,z \ge 1</math>.
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Now, we subtract to find <math>2y+3z=12</math>, and <math>y,z \ge 1</math>.
It follows that <math>y</math> is a multiple of <math>3</math> and <math>2y</math> is a multiple of <math>6</math>, so since <math>0<2y<12</math>, we must have <math>2y=6</math>.
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It follows that <math>2y</math> is a multiple of <math>3</math> and <math>3z</math> is a multiple of <math>3</math>. Since sum of 2 multiples of 3 = multiple of 3, so we must have <math>2y=6</math>.
  
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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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.
  
==Solution 2==
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===Solution 2===
 
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>.
 
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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<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
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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>.
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===Solution 3===
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Since there are 12 pairs of 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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===Solution 4===
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Let the amount of <math>1</math> dollar socks be <math>a</math>, <math>3</math> dollar socks be <math>b</math>, and <math>4</math> dollar socks be <math>c</math>. We then know that <math>a+b+c=12</math> and <math>a+3b+4c=24</math>. We can make <math>a+b+c=12</math> into <math>a=12-b-c</math> and then plug that into the other equation, producing <math>12-b-c+3b+4c=24</math> which simplifies to <math>2b+3c=12</math>. It's not hard to see <math>b=3</math> and <math>c=2</math>. Now that we know <math>b</math> and <math>c</math>, we know that <math>a=7</math>, meaning the number of <math>1</math> dollar socks Ralph bought is <math>\boxed{\textbf{(D)} 7}</math>.
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/toZI27nNHeQ
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 +
~Education, the Study of Everything
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===Video Solution===
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https://youtu.be/hvnVuLbveJs
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~savannahsolver
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==Video Solution by OmegaLearn==
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https://youtu.be/rQUwNC0gqdg?t=2187
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~pi_is_3.14
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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 00:06, 17 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 (HOW TO THINK CRITICALLY!!!)

https://youtu.be/toZI27nNHeQ

~Education, the Study of Everything


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
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All AJHSME/AMC 8 Problems and Solutions

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