Difference between revisions of "2015 AMC 8 Problems/Problem 20"
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<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== |
==See Also== | ==See Also== |
Revision as of 15:01, 12 September 2020
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?
Contents
Solution 1
So let there be pairs of socks, pairs of socks, pairs of 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 .
Solution 3
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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