Difference between revisions of "2015 AMC 8 Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
− | So let there be <math>x</math> pairs of <math>1</math> | + | So let there be <math>x</math> pairs of <math>\$1</math> socks, <math>y</math> pairs of <math>\$3</math> socks, <math>z</math> pairs of <math>\$4</math> socks. |
We have <math>x+y+z=12</math>, <math>x+3y+4z=24</math>, and <math>x,y,z \ge 1</math>. | We have <math>x+y+z=12</math>, <math>x+3y+4z=24</math>, and <math>x,y,z \ge 1</math>. |
Revision as of 14:13, 3 December 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
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
.
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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