Difference between revisions of "2019 AMC 10B Problems/Problem 11"
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Because <math>\frac{9x}{10}</math> is the number of blue marbles in Jar <math>1</math>, and <math>\frac{8x}{9}</math> is the number of blue marbles in Jar <math>2</math>, there are <math>\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5</math> more marbles in Jar <math>1</math> than Jar <math>2</math>. This means the answer is <math>\boxed{\textbf{(A) } 5}</math>. | Because <math>\frac{9x}{10}</math> is the number of blue marbles in Jar <math>1</math>, and <math>\frac{8x}{9}</math> is the number of blue marbles in Jar <math>2</math>, there are <math>\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5</math> more marbles in Jar <math>1</math> than Jar <math>2</math>. This means the answer is <math>\boxed{\textbf{(A) } 5}</math>. | ||
+ | |||
+ | ==Solution 2(Completely Solve)== | ||
+ | Let <math>b_1</math>, <math>g_1</math>, <math>b_2</math>, <math>g_2</math>, 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, | ||
+ | <math>\frac{</math>b_1<math>}{</math>g_1<math>} = \frac{9}{1}</math>, <math>\frac{</math>b_2<math>}{</math>g_2<math>} = \frac{8}{1}</math>, <math>g_1 + g_2 =95</math>, and <math>b_1 + g_1 = b_2 + g_2</math>. | ||
+ | Since <math>b_1 = 9g_1</math> and <math>b_2 = 8g_2</math>, we substitue that in to obtain <math>10g_1 = 9g_2</math>. | ||
+ | Coupled with our third equation, we find that <math>g_1 = 45</math>, and that <math>g_2 = 50</math>. We now use this information to find <math>b_1 = 405</math> | ||
+ | and <math>b_2 = 400</math>. Therefore, <math>b_1 - b_2 = 5</math> so our answer is <math>\boxed{\textbf{(A) } 5}</math>. | ||
+ | ~Binderclips1 | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:56, 30 May 2020
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
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,
$\frac{$ (Error compiling LaTeX. Unknown error_msg)b_1$}{$ (Error compiling LaTeX. Unknown error_msg)g_1$} = \frac{9}{1}$ (Error compiling LaTeX. Unknown error_msg), $\frac{$ (Error compiling LaTeX. Unknown error_msg)b_2$}{$ (Error compiling LaTeX. Unknown error_msg)g_2$} = \frac{8}{1}$ (Error compiling LaTeX. Unknown error_msg),
, and
.
Since
and
, we substitue 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
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.