Difference between revisions of "2015 AMC 8 Problems/Problem 20"
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| − | ==Solution== | + | ==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== | ||
| + | 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>. | ||
| + | |||
| + | There are two ways to make packages of socks that average to <math>\$2</math>. You can have: | ||
| + | |||
| + | <math>\bullet</math> Two <math>\$1</math> pairs and one <math>\$4</math> pair (package adds up to <math>\$6</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 | ||
| + | <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>. | ||
==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 00:59, 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?
Solution 1
So let there be 
pairs of 
dollar socks, 
pairs of 
dollar socks, 
pairs of 
dollar socks.
We have 
, 
, and 
.
Now we subtract to find 
, and 
.
It follows that is a multiple of and 
is a multiple of 
, so since 
, we must have 
.
Therefore, 
, and it follows that 
. Now 
, as desired.
Solution 2
Since the total cost of the socks was 
and Ralph bought 
pairs, the average cost of each pair of socks is 
.
There are two ways to make packages of socks that average to 
. You can have:
Two 
pairs and one 
pair (package adds up to 
)
One pair and one 
pair (package adds up to )
So now we need to solve
![\[6a+4b=24,\]](http://latex.artofproblemsolving.com/e/c/8/ec8c1f2366e495759813e9b411c05f8f5b3e0c50.png)
where 
is the number of packages and 
is the number of packages. We see our only solution (that has at least one of each pair of sock) is 
, which yields the answer of 
.
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. 
