AMC 10 · 2008 · #12

Grade 7 algebra

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Chained Ratio Linking

percentageratio-proportionformula-substitution convert-to-algebra ↑ Prerequisites: percentage

📏 Medium solution 💡 1 insight

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?

Pick an answer.

(A)

$\ 2.85r$

(B)

$\ 3r$

(C)

$\ 3.4r$

(D)

$\ 3.85r$

(E)

$\ 4.25r$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A collection has red, blue, and green marbles. There are 25% more red marbles than blue, and 60% more green marbles than red. Writing the count of red marbles as $r$, find the total number of marbles in terms of $r$.

Givens: Red is 25% more than blue.; Green is 60% more than red.; The number of red marbles is $r$.; Answer choices: (A) $2.85r$, (B) $3r$, (C) $3.4r$, (D) $3.85r$, (E) $4.25r$.

Unknowns: The total number of marbles, expressed as a multiple of $r$.

Understand

Givens: Red is 25% more than blue.; Green is 60% more than red.; The number of red marbles is $r$.; Answer choices: (A) $2.85r$, (B) $3r$, (C) $3.4r$, (D) $3.85r$, (E) $4.25r$.

Plan

Primary tool: #4 Introduce a Variable

Secondary: #13 Convert to Algebra, #8 Analyze the Units, #3 Eliminate Possibilities

Everything is measured against $r$, the red count, so I express blue and green as multiples of $r$ (Tool #4). The two percent phrases become equations (Tool #13); the trick is spotting which color is the base each time. Reading '% more' as 'base plus that percent of the base' (Tool #8) keeps blue and green straight. Adding the three multiples gives one coefficient, and matching it to the list (Tool #3) picks the answer.

Execute — Answer: C

#4 Introduce a Variable 7.RP.A.3 Step 1

Rewrite blue from the red count

'25% more red than blue' means red equals blue plus a quarter of blue, so red $= 1.25 \times$ blue.
Blue is the base here, not red.
To get blue from $r$, divide: blue $= r / 1.25 = 0.8r$.
So there are fewer blue marbles than red, which makes sense because red is the larger, increased amount.

$$r = 1.25 \cdot b \;\Rightarrow\; b = \frac{r}{1.25} = 0.8r$$

💡 'A is 25% more than B' makes B the base, so you divide by 1.25 to go from A back to B.

#13 Convert to Algebra 7.RP.A.3 Step 2

Rewrite green from the red count

'60% more green than red' means green equals red plus $60\%$ of red, and this time red is the base.
So green $= r + 0.6r = 1.6r$.
Green is the largest pile because it is red increased by more than half.

$$g = r + 0.6r = 1.6r$$

💡 'More than red' means red is the base, so add the percent directly to $r$.

#13 Convert to Algebra 6.EE.A.3 Step 3

Add the three amounts

The total is blue plus red plus green.
Substitute each as a multiple of $r$: $0.8r + r + 1.6r$.
These are like terms, so add the coefficients $0.8 + 1 + 1.6 = 3.4$.
The total is $3.4r$.

$$0.8r + r + 1.6r = (0.8 + 1 + 1.6)r = 3.4r$$

💡 Once every color is a number of $r$'s, adding the totals is just adding those numbers.

#3 Eliminate Possibilities 6.RP.A.3 Step 4

Match the coefficient to a choice

The total $3.4r$ appears exactly in the list of choices.
The other options come from mistakes like using $0.75r$ for blue instead of $0.8r$.
Since $3.4r$ matches, the answer is (C).

$$3.4r \;\Rightarrow\; \textbf{(C)}$$

💡 The right total should land on a listed value with no rounding, and $3.4r$ does.

[1] #4 7.RP.A.3 '25% more red than blue' means red equals blue plus a quarter of blue, so red $=

[2] #13 7.RP.A.3 '60% more green than red' means green equals red plus $60\%$ of red, and this ti

[3] #13 6.EE.A.3 The total is blue plus red plus green. Substitute each as a multiple of $r$: $0.

[4] #3 6.RP.A.3 The total $3.4r$ appears exactly in the list of choices. The other options come

Review

Reasonableness: Test with a concrete count. Let blue $= 100$. Then red is $25\%$ more: $r = 125$. Green is $60\%$ more than red: $1.6 \times 125 = 200$. The total is $100 + 125 + 200 = 425$. As a multiple of $r$, that is $425 / 125 = 3.4$, so the total is $3.4r$. This confirms (C). Note the total is a bit more than $3r$ because green is the biggest group, which fits.

Alternative: Set $r = 100$ directly. Blue $= 100 / 1.25 = 80$ and green $= 1.6 \times 100 = 160$. Total $= 80 + 100 + 160 = 340 = 3.4 \times 100 = 3.4r$, the same choice (C).

CCSS standards used (min grade 7)

7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems (Reading '25% more' and '60% more' as percent increases and rewriting blue as $0.8r$ and green as $1.6r$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Combining the like terms $0.8r + r + 1.6r$ into the single expression $3.4r$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Checking the coefficient against the answer choices and verifying with the concrete counts $100, 125, 200$.)

⭐ In '25% more A than B', B is the base you start from, so blue is $r/1.25 = 0.8r$; add blue, red, and green all measured in $r$'s and you get $3.4r$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Rewrite blue from the red count

Rewrite green from the red count

Add the three amounts

Match the coefficient to a choice

Review

CCSS standards used (min grade 7)

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