Difference between revisions of "2008 AMC 10A Problems/Problem 12"
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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>. | |
+ | |||
+ | ==Solution 2== | ||
+ | Let the number of red marbles be 100. You have <math>b = 80,</math> and <math>g = 160.</math> | ||
+ | |||
+ | The answer is <math>\frac{100+80+160}{100} = \boxed{3.4}</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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:26, 28 September 2021
Contents
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 .
Solution 2
Let the number of red marbles be 100. You have 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 |
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