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Difference between revisions of "2008 AMC 10A Problems/Problem 12"

(Added problem, solution still needed)
 
(Solution + Typo in Problem)
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==Problem==
 
==Problem==
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In a collection of red, blue, and green marbles, there are <math>25\%</math> more red marbles than blue marbles, and there are <math>60\%</math> mroe green marbles than red marbles. Suppose that there are <math>r</math> red marbles. What is the total number of marbles in the collection?
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In a collection of red, blue, and green marbles, there are <math>25\%</math> more red marbles than blue marbles, and there are <math>60\%</math> more green marbles than red marbles. Suppose that there are <math>r</math> red marbles. What is the total number of marbles in the collection?
  
 
<math>\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r</math>
 
<math>\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r</math>
  
 
==Solution==
 
==Solution==
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{{solution}}
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The number of blue marbles is <math>\frac{4}{5}r</math>, the number of green marbles is <math>\frac{8}{5}r</math>, and the number of red marbles is <math>r</math>.
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The total number of marbles is <math>\frac{4}{5}r+\frac{8}{5}r+r=3.4r</math>. Thus, the answer is <math>\mathrm{(C)}</math>.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2008|ab=A|num-b=11|num-a=13}}

Revision as of 00:14, 26 April 2008

Problem

In a collection of red, blue, and green marbles, there are $25\%$ more red marbles than blue marbles, and there are $60\%$ more green marbles than red marbles. Suppose that there are $r$ red marbles. What is the total number of marbles in the collection?

$\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r$

Solution

The number of blue marbles is $\frac{4}{5}r$, the number of green marbles is $\frac{8}{5}r$, and the number of red marbles is $r$.

The total number of marbles is $\frac{4}{5}r+\frac{8}{5}r+r=3.4r$. Thus, the answer is $\mathrm{(C)}$.

See also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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