Difference between revisions of "2019 AMC 10B Problems/Problem 11"
(→Solutions) |
|||
(32 intermediate revisions by 21 users not shown) | |||
Line 1: | Line 1: | ||
− | + | ==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> | ||
+ | |||
+ | ==Solutions== | ||
+ | |||
+ | |||
+ | ===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== | ||
+ | |||
+ | {{AMC10 box|year=2019|ab=B|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:56, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.