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2008 AMC 10A Problem 12

2008 AMC 10A Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P11-15

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?

A)
$2.85r$
B)
$3r$
C)
$3.4r$
D)
$3.85r$
E)
$4.25r$

Answer

Key insight

Red is 1.25 times blue, so blue is r/1.25 = 0.8r; green is 1.6r; the total is (0.8 + 1 + 1.6)r = 3.4r.

Solution

Let $b$ and $g$ be the numbers of blue and green marbles.

" $25\%$ more red than blue" means $r = 1.25b$ , so
$b = \frac{r}{1.25} = \frac{4r}{5} = 0.8r.$

" $60\%$ more green than red" means $g = 1.6r$ .

The total is
$r + b + g = r + 0.8r + 1.6r = 3.4r.$

The answer is $\boxed{\textbf{(C)}\ 3.4r}$ .

Why this works

Each "percent more than" statement is a multiplier applied to the base quantity named after "than." When the unknown is the larger quantity, undo the increase by dividing, not by subtracting the same percent: $25\%$ more than $b$ is $r$ , so $b$ is $r$ divided by $1.25$ , which is $20\%$ less than $r$ , not $25\%$ less.

Alternative approach

Pick numbers: let $b = 4$ , so $r = 5$ and $g = 8$ . The total is $17 = 3.4 \cdot 5$ .

The trap

Taking blue to be 25% less than red, 0.75r, instead of r/1.25 = 0.8r, which gives 3.35r and a guess at (B) or (C).

Common mistakes

Taking blue to be 25% less than red, 0.75r, instead of r/1.25 = 0.8r, which gives 3.35r and a guess at (B) or (C).
Applying $60\%$ to blue instead of red, or adding percents ( $25\% + 60\%$ ) as if they had the same base.

Techniques

Set up the equation/formula and compute; no special trick needed