2008 AMC 10A Problems/Problem 12

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 $\mathrm{(C)}$ .

Solution 2

Let the number of red marbles be 100. You have $b = 80,$ and $g = 160.$

The answer is $\frac{100+80+160}{100} = \boxed{3.4}$

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

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2008 AMC 10A Problems/Problem 12

Contents

Problem

Solution

Solution 2

See also