2008 AMC 10A Problems/Problem 12

Revision as of 00:15, 26 April 2008 by I like pie (talk | contribs) (Slight change in wording of solution)

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 $r$.

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

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 10 Problems and Solutions