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Difference between revisions of "2019 AMC 10B Problems/Problem 11"

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(Added solution)
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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>.
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 +
==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,
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<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>.
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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>.
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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>
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and <math>b_2 = 400</math>. Therefore, <math>b_1 - b_2 = 5</math> so our answer is <math>\boxed{\textbf{(A) } 5}</math>.
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~Binderclips1
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:56, 29 May 2020

Problem

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$, and the ratio of blue to green marbles in Jar $2$ is $8:1$. There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$?

$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$

Solution

Call the number of marbles in each jar $x$ (because the problem specifies that they each contain the same number). Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$, and $\frac{x}{9}$ is the number of green marbles in Jar $2$. Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$, we have $\frac{19x}{90}=95$, so there are $x=450$ marbles in each jar.

Because $\frac{9x}{10}$ is the number of blue marbles in Jar $1$, and $\frac{8x}{9}$ is the number of blue marbles in Jar $2$, there are $\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5$ more marbles in Jar $1$ than Jar $2$. This means the answer is $\boxed{\textbf{(A) } 5}$.

Solution 2(Completely Solve)

Let $b_1$, $g_1$, $b_2$, $g_2$, 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), $g_1 + g_2 =95$, and $b_1 + g_1 = b_2 + g_2$. Since $b_1 = 9g_1$ and $b_2 = 8g_2$, we substitue that in to obtain $10g_1 = 9g_2$. Coupled with our third equation, we find that $g_1 = 45$, and that $g_2 = 50$. We now use this information to find $b_1 = 405$ 

and $b_2 = 400$. Therefore, $b_1 - b_2 = 5$ so our answer is $\boxed{\textbf{(A) } 5}$.

~Binderclips1

Video Solution

https://youtu.be/mXvetCMMzpU

~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

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