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 01:59, 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 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 |
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