Difference between revisions of "2019 AMC 10B Problems/Problem 11"
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− | + | ==Problem== | |
+ | |||
+ | Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar <math>1</math> the ratio of blue to green marbles is <math>9:1</math>, and the ratio of blue to green marbles in Jar <math>2</math> is <math>8:1</math>. There are <math>95</math> green marbles in all. How many more blue marbles are in Jar <math>1</math> than in Jar <math>2</math>? | ||
+ | |||
+ | <math>\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50</math> | ||
+ | |||
+ | ==Solution== | ||
+ | Call the number of marbles in each jar <math>x</math> (because the problem specifies that they each contain the same number). Thus, <math>\frac{x}{10}</math> is the number of green marbles in Jar <math>1</math>, and <math>\frac{x}{9}</math> is the number of green marbles in Jar <math>2</math>. Since <math>\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}</math>, we have <math>\frac{19x}{90}=95</math>, so there are <math>x=450</math> marbles in each jar. | ||
+ | |||
+ | 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{b_1}{g_1} = \frac{9}{1}</math>, | ||
+ | <math>\frac{b_2}{g_2} = \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 | ||
+ | |||
+ | ~LaTeX fixed by Starshooter11 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/mXvetCMMzpU | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:40, 8 June 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, , , , 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
~LaTeX fixed by Starshooter11
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.