Difference between revisions of "2019 AMC 10B Problems/Problem 11"
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+ | ==Solution 3== | ||
+ | Call the number of green marbles in Jar 1 <math>x</math>, and the number of green marbles in Jar 2 <math>y</math>. | ||
+ | Realize that this is the Diophantine equation <math>10x+9y=95</math> where the only solution we care about is the integer one. Immediately realize that since <math>10x</math> will end with a <math>0</math> no matter what, <math>9y</math> must end in a <math>5</math>. The only <math>y</math> that works is <math>y=5</math> which leads to <math>x=5</math> and we find that there are <math>45</math> blue marbles in Jar 1, and <math>40</math> in Jar 2. The difference is <math>45-40=5</math>. | ||
+ | ~JH. L | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:27, 9 June 2022
Contents
Problem
Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar the ratio of blue to green marbles is
, and the ratio of blue to green marbles in Jar
is
. There are
green marbles in all. How many more blue marbles are in Jar
than in Jar
?
Solution 1
Call the number of marbles in each jar (because the problem specifies that they each contain the same number). Thus,
is the number of green marbles in Jar
, and
is the number of green marbles in Jar
. Since
, we have
, so there are
marbles in each jar.
Because is the number of blue marbles in Jar
, and
is the number of blue marbles in Jar
, there are
more marbles in Jar
than Jar
. This means the answer is
.
Solution 2 (Completely Solve)
Let ,
,
,
, represent the amount of blue marbles in jar 1, the amount of green marbles in jar 1, the
the amount of blue marbles in jar 2, and the amount of green marbles in jar 2, respectively. We now have the equations,
,
,
, and
.
Since
and
, we substitute that in to obtain
.
Coupled with our third equation, we find that
, and that
. We now use this information to find
and
.
Therefore, so our answer is
.
~Binderclips1
~LaTeX fixed by Starshooter11 ~Typo fixed by Little
Solution 3
Call the number of green marbles in Jar 1 , and the number of green marbles in Jar 2
.
Realize that this is the Diophantine equation
where the only solution we care about is the integer one. Immediately realize that since
will end with a
no matter what,
must end in a
. The only
that works is
which leads to
and we find that there are
blue marbles in Jar 1, and
in Jar 2. The difference is
.
~JH. L
Video Solution
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.