Difference between revisions of "2015 AMC 8 Problems/Problem 20"
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<math>\bullet</math> One <math>\$1</math> pair and one <math>\$3</math> pair (package adds up to <math>\$4</math>) | <math>\bullet</math> One <math>\$1</math> pair and one <math>\$3</math> pair (package adds up to <math>\$4</math>) | ||
− | + | Now, we need to solve | |
<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>. |
Revision as of 06:34, 4 May 2023
Contents
Problem
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?
Solutions
Solution 1
So, let there be pairs of
socks,
pairs of
socks, and
pairs of
socks.
We have ,
, and
.
Now, we subtract to find , and
.
It follows that
is a multiple of
and
is a multiple of
. Since sum of 2 multiples of 3 = multiple of 3, so 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
)
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
Since there are 12 pairs of 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.
Solution 4
Let the amount of dollar socks be
,
dollar socks be
, and
dollar socks be
. We then know that
and
. We can make
into
and then plug that into the other equation, producing
which simplifies to
. It's not hard to see
and
. Now that we know
and
, we know that
, meaning the number of
dollar socks Ralph bought is
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=2187
~pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.