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

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==Solution 3==
 
==Solution 3==
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Writing out to ratios, we have <math>9:1</math> in jar <math>1</math> and <math>8:1</math> in jar <math>2</math>. Since the jar must have to same amount of marbles, let's make a variable <math>a</math> and <math>b</math> for each of the ratios to be multiplied by. Now we would have <math>9a + a = 8b + b \rightarrow 10a = 9b</math>. Because there are possibly
+
Writing out to ratios, we have <math>9:1</math> in jar <math>1</math> and <math>8:1</math> in jar <math>2</math>. Since the jar must have to same amount of marbles, let's make a variable <math>a</math> and <math>b</math> for each of the ratios to be multiplied by. Now we would have <math>9a + a = 8b + b \rightarrow 10a = 9b</math>. Because this a diophantine equation, let's start from use the most basic <math>a</math> and <math>b</math> that works and scale it from there. We should be able to see that <math>a</math> and <math>b</math> could be <math>9</math> and <math>10</math> respectively. Now remember that there are <math>95</math> green marbles or <math>x(a + b) = 95</math> for some integer <math>x</math> to scale it. Substituting and dividing, we find <math>x = 5</math>. Thus to find the difference of the blue marbles we must do
 +
<cmath>\begin{align*}
 +
x(9a - 8b) &= \\
 +
5(81 - 80) &= \\
 +
5(1) &= 5 \\
 +
\end{align*}</cmath>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 07:52, 13 October 2023

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 1

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, $\frac{b_1}{g_1} = \frac{9}{1}$, $\frac{b_2}{g_2} = \frac{8}{1}$, $g_1 + g_2 =95$, and $b_1 + g_1 = b_2 + g_2$. Since $b_1 = 9g_1$ and $b_2 = 8g_2$, we substitute 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

~LaTeX fixed by Starshooter11 ~Typo fixed by Little

Solution 3

Writing out to ratios, we have $9:1$ in jar $1$ and $8:1$ in jar $2$. Since the jar must have to same amount of marbles, let's make a variable $a$ and $b$ for each of the ratios to be multiplied by. Now we would have $9a + a = 8b + b \rightarrow 10a = 9b$. Because this a diophantine equation, let's start from use the most basic $a$ and $b$ that works and scale it from there. We should be able to see that $a$ and $b$ could be $9$ and $10$ respectively. Now remember that there are $95$ green marbles or $x(a + b) = 95$ for some integer $x$ to scale it. Substituting and dividing, we find $x = 5$. Thus to find the difference of the blue marbles we must do \begin{align*} x(9a - 8b) &= \\ 5(81 - 80) &= \\ 5(1) &= 5 \\ \end{align*}

Video Solution

https://youtu.be/gQRZqfZBUAY

~Education, the Study of Everything

Video Solution

https://youtu.be/DzQZtQvNDwA?t=9

Video Solution

https://youtu.be/mXvetCMMzpU

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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