Difference between revisions of "2012 AMC 8 Problems/Problem 19"
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<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12 </math> | <math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12 </math> | ||
− | ==Solution== | + | ==Solution 1 (Trial and Error)== |
− | + | <math>6</math> are blue and green - <math>b+g=6</math> | |
− | + | <math>8</math> are red and blue - <math>r+b=8</math> | |
− | <math> g | + | <math>4</math> are red and green- <math>r+g=4</math> |
− | |||
− | <math> r + g = 4 </math> | + | We can do trial and error. Let's make blue <math>5</math>. That makes green <math>1</math> and red <math>3</math> because <math>6-5=1</math> and <math>8-5=3</math>. To check this, let's plug <math>1</math> and <math>3</math> into <math>r+g=4</math>, which works. Now count the number of marbles - <math>5+3+1=9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}.</math> |
− | + | ==Solution 2== | |
− | We add all the | + | We already knew the facts: <math>6</math> are blue and green, meaning <math>b+g=6</math>; <math>8</math> are red and blue, meaning <math>r+b=8</math>; <math>4</math> are red and green, meaning <math>r+g=4</math>. Then we need to add these three equations: <math>b+g+r+b+r+g=2(r+g+b)=6+8+4=18</math>. It gives us all of the marbles are <math>r+g+b = 18/2 = 9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. |
+ | ~LarryFlora | ||
− | + | ==Solution 3 (Venn Diagrams)== | |
+ | We may draw three Venn diagrams to represent these three cases, respectively. | ||
− | + | [[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]] | |
− | <math> | + | Let the amount of all the marbles be <math>x</math>, meaning <math>R+G+B = x</math>. |
+ | |||
+ | The Venn diagrams give us the equation: <math>x = (x-6)+(x-8)+(x-4)</math>. | ||
+ | So <math>x = 3x-18</math>, <math>x = 18/2 =9</math>. | ||
+ | Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. | ||
+ | ~LarryFlora | ||
+ | |||
+ | ==Solution 4 (Answer Choices)== | ||
+ | Since we know all but <math>8</math> marbles in the jar are green, the jar must have at least <math>9</math> marbles. Then we can just start from <math>C</math> and keep going. If there are <math>9</math> marbles total, there are <math>3</math> red marbles <math>(9-6)</math>, <math>1</math> green marble <math>(9-8)</math>, and <math>5</math> blue marbles <math>(9-4)</math>. Since we assumed there were <math>9</math> marbles and <math>3+1+5=9</math>, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. | ||
+ | |||
+ | ==Solution 5 (Algebra) == | ||
+ | Let <math>x</math> be the number of total marbles. There are <math>x – 6</math> red marbles, <math>x – 8</math> green marbles, and <math>x – 4</math> blue marbles. | ||
+ | We can create an equation: <math>(x – 6)+(x – 8)+(x – 4)=x</math> | ||
+ | Solving, we get <math>x=9</math>, which means the total number of marbles is <math>\boxed{\textbf{(C)}\ 9}</math>. | ||
+ | -J.L.L | ||
+ | (Feel free to edit) | ||
+ | |||
+ | ==Solution 6== | ||
+ | Let <math>x</math> be the number of total marbles, <math>r</math> be the number of red marbles, <math>g</math> be the number of green marbles, and <math>b</math> be the number of blue marbles. Then we have <math>x - r = 6</math>, <math>x - g = 8</math>, <math>x - b = 4</math>, and <math>r + g + b = x</math>. Adding the first three equations together, we get <math>3x - r - g - b = 18</math> or <math>3x - (r + g + b) = 18</math>. Substituting in the fourth equation, we have <math>3x - x = 18</math> <math>\implies</math> <math>\boxed{\textbf{(C)}\ 9}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/mMph7QH1kX0 Soo, DRMS, NM | ||
+ | |||
+ | https://youtu.be/-p5qv7DftrU ~savannahsolver | ||
+ | |||
+ | ==Video Solution by OmegaLearn == | ||
+ | https://youtu.be/TkZvMa30Juo?t=1316 | ||
+ | |||
+ | ~pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=18|num-a=20}} | {{AMC8 box|year=2012|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:30, 18 February 2024
Contents
Problem
In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
Solution 1 (Trial and Error)
are blue and green -
are red and blue -
are red and green-
We can do trial and error. Let's make blue . That makes green
and red
because
and
. To check this, let's plug
and
into
, which works. Now count the number of marbles -
. So the answer is
Solution 2
We already knew the facts: are blue and green, meaning
;
are red and blue, meaning
;
are red and green, meaning
. Then we need to add these three equations:
. It gives us all of the marbles are
. So the answer is
.
~LarryFlora
Solution 3 (Venn Diagrams)
We may draw three Venn diagrams to represent these three cases, respectively.
Let the amount of all the marbles be , meaning
.
The Venn diagrams give us the equation: .
So
,
.
Thus, the answer is
.
~LarryFlora
Solution 4 (Answer Choices)
Since we know all but marbles in the jar are green, the jar must have at least
marbles. Then we can just start from
and keep going. If there are
marbles total, there are
red marbles
,
green marble
, and
blue marbles
. Since we assumed there were
marbles and
, the answer is
.
Solution 5 (Algebra)
Let be the number of total marbles. There are
red marbles,
green marbles, and
blue marbles.
We can create an equation:
Solving, we get
, which means the total number of marbles is
.
-J.L.L
(Feel free to edit)
Solution 6
Let be the number of total marbles,
be the number of red marbles,
be the number of green marbles, and
be the number of blue marbles. Then we have
,
,
, and
. Adding the first three equations together, we get
or
. Substituting in the fourth equation, we have
.
Video Solution
https://youtu.be/mMph7QH1kX0 Soo, DRMS, NM
https://youtu.be/-p5qv7DftrU ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=1316
~pi_is_3.14
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.