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>. | ||
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+ | ==Solution 3 -SweetMango77== | ||
+ | Since there are 12 pairs 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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==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 16:33, 28 October 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?
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 -SweetMango77
Since there are 12 pairs socks, and Ralph bought at least one pair of each, there are pairs of socks left. Also, the sum of the three pairs of socks is . This means that there are dollars left. If there are only dollar socks left, then we would have dollars wasted, which leaves more dollars. If we replace one pair with a dollar pair, then we would waste an additional dollars. If we replace one pair with a dollar pair, then we would waste an additional dollars. The only way can be represented as a sum of s and s is . If we change pairs, we would have pairs left. Adding the one pair from previously, we have pairs.
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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