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

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(→‎Solution)
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==Solution==
 
==Solution==
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Let <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.
  
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Let <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.
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If "all but 6 are red marbles", that means that the number of green marbles and the number of blue marbles amount to <math>6</math>. Likewise, the number of red marbles and blue marbles amount to <math>8</math>, and the number of red marbles and the number of green marbles amount to <math>4</math>.
  
 
We have three equations:
 
We have three equations:
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<math> r + g = 4 </math>
 
<math> r + g = 4 </math>
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Now we use some algebraic manipulation.
 
  
 
We add all the equations to obtain a fourth equation:
 
We add all the equations to obtain a fourth equation:
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Now divide by <math> 2 </math> on both sides to find the total number of marbles:
 
Now divide by <math> 2 </math> on both sides to find the total number of marbles:
  
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<math> r + g + b = 9 </math>. The total number of marbles in the jar is <math> \boxed{\textbf{(C)}\ 9} </math>.
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<math> r + g + b = 9 </math>
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 +
Since the sum of the number of red marbles, green marbles, and blue marbles is the number of marbles in the jar, the total number of marbles in the jar is <math> \boxed{\textbf{(C)}\ 9} </math>.
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Notice how we never knew how many of each color there were (there is 1 green marble, 5 blue marbles, and 3 red marbles).
  
 
==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}}

Revision as of 12:15, 25 November 2013

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

Let $r$ be the number of red marbles, $g$ be the number of green marbles, and $b$ be the number of blue marbles.

If "all but 6 are red marbles", that means that the number of green marbles and the number of blue marbles amount to $6$. Likewise, the number of red marbles and blue marbles amount to $8$, and the number of red marbles and the number of green marbles amount to $4$.

We have three equations:

$g + b = 6$

$r + b = 8$

$r + g = 4$

We add all the equations to obtain a fourth equation:

$2r + 2g + 2b = 18$

Now divide by $2$ on both sides to find the total number of marbles:

$r + g + b = 9$

Since the sum of the number of red marbles, green marbles, and blue marbles is the number of marbles in the jar, the total number of marbles in the jar is $\boxed{\textbf{(C)}\ 9}$.

Notice how we never knew how many of each color there were (there is 1 green marble, 5 blue marbles, and 3 red marbles).

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

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