Difference between revisions of "2012 AMC 8 Problems/Problem 19"

Revision as of 18:14, 8 August 2023

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?

$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$

Solution 1

6 are blue and green- b+g=6

8 are red and blue- r+b=8

4 are red and green- r+g=4

We can do trial and error. Let's make blue 5. That makes green 1 and red 3 because 6-5=1 and 8-5=3. To check this let's plug 1 and 3 into r+g=4 and it does work. Now count the number of marbles- 5+3+1=9. So 9 (C) is the answer.

Solution 2

We already knew the facts: $6$ are blue and green, meaning $b+g=6$ ; $8$ are red and blue, meaning $r+b=8$ ; $4$ are red and green, meaning $r+g=4$ . Then we need to add these three equations: $b+g+r+b+r+g=2(r+g+b)=6+8+4=18$ . It gives us all of the marbles are $r+g+b = 18/2 = 9$ . So the answer is $\boxed{\textbf{(C)}\ 9}$ . ---LarryFlora

Solution 3 Venn Diagrams

We may draw three Venn diagrams to represent these three cases, respectively.

Let the amount of all the marbles is $x$ , meaning $R+G+B = x$ .

The Venn diagrams give us the equation: $x = (x-6)+(x-8)+(x-4)$ . So $x = 3x-18$ , $x = 18/2 =9$ . Thus, the answer is $\boxed{\textbf{(C)}\ 9}$ . ---LarryFlora

Solution 4 Venn Diagrams

We may draw three Venn diagrams to represent these three cases, respectively.

Let the amount of all the marbles is $x$ , meaning $R+G+B = x$ .

Adding the three Venn diagrams, it gives us the equation: $x+18 = 3x$ . So $2x = 18$ , $x = 18/2 =9$ . Thus, the answer is $\boxed{\textbf{(C)}\ 9}$ . ---LarryFlora

Solution 5 (Answer Choices)

Since we know all but $8$ marbles in the jar are green, the jar must have at least $9$ marbles. Then we can just start from $C$ and keep going. If there are $9$ marbles total, there are $3$ red marbles $(9-6)$ , $1$ green marble $(9-8)$ , and $5$ blue marbles $(9-4)$ . Since we assumed there were $9$ marbles and $3+1+5=9$ , the answer is $\boxed{\textbf{(C)}\ 9}$ .

Solution 6

Let $x$ be the number of total marbles. There are $x – 6$ red marbles, $x – 8$ green marbles, and $x – 4$ blue marbles. We can create an equation: $(x – 6)+(x – 8)+(x – 4)=x$ Solving, we get $x=9$ , which means the total number of marbles is $\boxed{\textbf{(C)}\9$ (Error compiling LaTeX. Unknown error_msg).

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

@@ Line 49: / Line 49: @@
 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>.
+Solving, we get <math>x=9</math>, which means the total number of marbles is <math>\boxed{\textbf{(C)}\9</math>.
 ==Video Solution==

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

Page

Toolbox

Search