Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3 -SweetMango77)
 
(21 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
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?
+
==Problem==
 +
 
 +
Ralph went to the store and bought 12 pairs of socks for a total of <math>\$24</math>. Some of the socks he bought cost <math>\$1</math> a pair, some of the socks he bought cost <math>\$3</math> a pair, and some of the socks he bought cost <math>\$4</math> a pair. If he bought at least one pair of each type, how many pairs of <math>\$1</math> socks did Ralph buy?
  
 
<math>
 
<math>
Line 9: Line 11:
 
</math>
 
</math>
  
−
==Solution 1==
+
==Solutions==
−
So let there be <math>x</math> pairs of <math>\$1</math> socks, <math>y</math> pairs of <math>\$3</math> socks, <math>z</math> pairs of <math>\$4</math> socks.
+
 
 +
===Solution 1===
 +
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>.
+
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>.
+
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.
+
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==
+
===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>.
  
Line 28: Line 32:
 
<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>.
  
−
==Solution 3 ==
+
===Solution 3===
−
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.
+
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.
 +
 
 +
===Solution 4===
 +
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>.
 +
 
 +
===Solution 5 (Guess and check)===
 +
If Ralph bought one sock of each kind, he already used <math>\$8</math>, so there are <math>\$16</math> left and 9 socks. If we split the <math>\$16</math> into four <math>\$4</math> sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a <math>\$3</math> pair, he would need to buy a <math>\$1</math> pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a <math>\$1</math> pair, he would either need to buy three <math>\$1</math> pairs or a <math>\$3</math> pair. If Ralph bought a <math>\$4</math> pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that <math>1 + 2 + 2 + 4 = 9</math>, or a <math>\$4</math> pair, two <math>\$3</math> pairs, and six <math>\$1</math> pairs. Because we subtracted the necessary one of each kind, there are two <math>\$4</math> pairs, three <math>\$3</math> pairs, and seven <math>\$1</math> pairs. Therefore, the number of <math>\$1</math> pairs Ralph bought is <math>\boxed{\textbf{(D)}~7}</math>.
 +
~strongstephen
 +
 
 +
== Video Solution by Pi Academy (Fast and Easy) ==
 +
 
 +
https://youtu.be/ugbi97qGScY?si=t3zYt83JEwa0PXtJ
 +
 
 +
~ jj_empire
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/toZI27nNHeQ
 +
 
 +
~Education, the Study of Everything
 +
 
 +
===Video Solution===
 +
https://youtu.be/hvnVuLbveJs
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/rQUwNC0gqdg?t=2187
 +
 
 +
~pi_is_3.14
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
  
 
==See Also==
 
==See Also==

Latest revision as of 17:55, 23 October 2024

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 $\boxed{\textbf{(D)} 7}$.

Solution 5 (Guess and check)

If Ralph bought one sock of each kind, he already used $$8$, so there are $$16$ left and 9 socks. If we split the $$16$ into four $$4$ sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a $$3$ pair, he would need to buy a $$1$ pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a $$1$ pair, he would either need to buy three $$1$ pairs or a $$3$ pair. If Ralph bought a $$4$ pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that $1 + 2 + 2 + 4 = 9$, or a $$4$ pair, two $$3$ pairs, and six $$1$ pairs. Because we subtracted the necessary one of each kind, there are two $$4$ pairs, three $$3$ pairs, and seven $$1$ pairs. Therefore, the number of $$1$ pairs Ralph bought is $\boxed{\textbf{(D)}~7}$. ~strongstephen

Video Solution by Pi Academy (Fast and Easy)

https://youtu.be/ugbi97qGScY?si=t3zYt83JEwa0PXtJ

~ jj_empire

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/toZI27nNHeQ

~Education, the Study of Everything

Video Solution

https://youtu.be/hvnVuLbveJs

~savannahsolver

Video Solution

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

~pi_is_3.14

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_20&oldid=230453"