Difference between revisions of "2019 AMC 10B Problems/Problem 11"
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==Problem== | ==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? | + | 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> | <math>\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50</math> | ||
− | ==Solution== | + | ==Solutions== |
− | Call the | + | |
+ | |||
+ | ===Solution 1=== | ||
+ | Our ratios are <math>9:1</math> in \( J_1 \) and <math>8:1</math> in \( J_2 \). | ||
+ | |||
+ | We start with the equation representing the total number of marbles in both jars, where \( x \) is the common multiplier: | ||
+ | |||
+ | <cmath> 9x + x + 8x + x = 19x </cmath> | ||
+ | |||
+ | |||
+ | Given that the total number of green marbles is 95: | ||
+ | |||
+ | <cmath> 19x = 95 </cmath> | ||
+ | |||
+ | Solving for \( x \): | ||
+ | |||
+ | <cmath> x = \frac{95}{19} = 5 </cmath> | ||
+ | |||
+ | The difference between blue marbles in \( J_1 \) and \( J_2 \) is simply | ||
+ | |||
+ | <math>9x-8x=9(5)-8(5)=45-40= </math><math>\boxed{\textbf{(A) } 5}</math>. | ||
+ | |||
+ | ~ GeometryMystery | ||
+ | |||
+ | |||
+ | ===Solution 2=== | ||
+ | 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 3 (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 substitute 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 | ||
+ | ~Typo fixed by Little | ||
+ | |||
+ | ===Solution 4=== | ||
+ | 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>. We can take the most obvious values of <math>a</math> and <math>b</math> and then 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) &= \boxed{\textbf{(B) }5} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~ Wiselion | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==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== | ==See Also== |
Latest revision as of 11:56, 21 June 2024
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 ?
Solutions
Solution 1
Our ratios are in \( J_1 \) and in \( J_2 \).
We start with the equation representing the total number of marbles in both jars, where \( x \) is the common multiplier:
Given that the total number of green marbles is 95:
Solving for \( x \):
The difference between blue marbles in \( J_1 \) and \( J_2 \) is simply
.
~ GeometryMystery
Solution 2
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 3 (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 4
Writing out to ratios, we have in jar and in jar . Since the jar must have to same amount of marbles, let's make a variable and for each of the ratios to be multiplied by. Now we would have . We can take the most obvious values of and and then scale it from there. We should be able to see that and could be and respectively. Now remember that there are green marbles or for some integer to scale it. Substituting and dividing, we find . Thus to find the difference of the blue marbles we must do
~ Wiselion
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/DzQZtQvNDwA?t=9
Video Solution
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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