2008 AMC 10A Problems/Problem 12: Difference between revisions
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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>. | 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>. | ||
Thus, the total number of marbles is <math>\frac{4}{5}r+\frac{8}{5}r+r=3.4r</math>, and 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 00:15, 26 April 2008
Problem
In a collection of red, blue, and green marbles, there are 
more red marbles than blue marbles, and there are 
more green marbles than red marbles. Suppose that there are 
red marbles. What is the total number of marbles in the collection?
Solution
The number of blue marbles is 
, the number of green marbles is 
, and the number of red marbles is .
Thus, the total number of marbles is 
, and the answer is 
.
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 | ||
