2008 AMC 10A Problems/Problem 12: Difference between revisions
No edit summary |
Mathboy282 (talk | contribs) |
||
| Line 8: | Line 8: | ||
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>. | 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}} | {{MAA Notice}} | ||
Latest revision as of 19:26, 28 September 2021
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
