2015 AMC 8

Complete problem set with solutions and individual problem pages

Problem 20 Hard

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?

  • A.

    4

  • B.

    5

  • C.

    6

  • D.

    7

  • E.

    8

Answer:D

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 \geqslant 1.

Now, we subtract to find 2y+3z=12, and y,z \geqslant 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}.

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