AMC 8 · 2007 · #17

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition

percentageratio-proportionfraction-arithmetic identify-subproblems ↑ Prerequisites: percentagefraction-arithmetic

📏 Short solution 💡 2 insights

Problem

A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow
tint and $45\%$ water. Five liters of yellow tint are added to
the original mixture. What is the percent of yellow tint
in the new mixture?

$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A $30$-liter paint mixture is $25\%$ red tint, $30\%$ yellow tint, and $45\%$ water. Five more liters of pure yellow tint are stirred in. What percent of the new mixture is yellow tint?

Givens: Original mixture: $30$ liters total; Original composition: $25\%$ red tint, $30\%$ yellow tint, $45\%$ water; $5$ liters of pure yellow tint are added; Answer choices: (A) $25$, (B) $35$, (C) $40$, (D) $45$, (E) $50$

Unknowns: The percent of yellow tint in the new mixture

Understand

Restated: A $30$-liter paint mixture is $25\%$ red tint, $30\%$ yellow tint, and $45\%$ water. Five more liters of pure yellow tint are stirred in. What percent of the new mixture is yellow tint?

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

The trap in this problem is reasoning with the percent $30\%$ directly — when you add liquid to one part, both the part and the whole change, so the percent does not just go up by some neat number. Tool #7 (Identify Subproblems) handles this by splitting the work into three independent steps: (a) convert the $30\%$ to actual liters of yellow, (b) update the liters of yellow and the total liters after the pour, and (c) convert back to a percent. Each step is one arithmetic move. Tool #1 (Draw a Diagram) supports this with a part-whole bar that makes the "yellow piece grows, total bar grows" picture explicit. We deliberately avoid Tool #13 (Algebra) because no equation is needed — just three lines of arithmetic on amounts.

Execute — Answer: C

#7 Identify Subproblems 6.RP.A.3 Step 1

Convert the starting percent to liters of yellow.
Of the $30$ liters of original mixture, $30\%$ is yellow tint, so use the "percent of" calculation.

$$\text{yellow}_{\text{old}} = 30\% \times 30 = 0.30 \times 30 = 9 \text{ liters}$$

💡 $30\%$ of $30$ is the same as $\tfrac{3}{10}$ of $30$, which is $9$ — a straight Grade 6 "percent of a quantity" move.

#1 Draw a Diagram 4.OA.A.3 Step 2

Update the yellow amount after pouring in $5$ more liters of pure yellow tint.
Only the yellow changes; red and water stay put.

$$\text{yellow}_{\text{new}} = 9 + 5 = 14 \text{ liters}$$

💡 On a part-whole bar, the yellow stripe stretches by $5$ liters while the red and water stripes do not move.

#1 Draw a Diagram 4.OA.A.3 Step 3

Update the total volume.
The added $5$ liters bumps the total mixture from $30$ liters up to $35$ liters.

$$\text{total}_{\text{new}} = 30 + 5 = 35 \text{ liters}$$

💡 The whole bar gets longer by exactly the $5$ liters that were poured in — nothing left the container.

#7 Identify Subproblems 6.RP.A.3 Step 4

Convert back to a percent.
Divide the new yellow amount by the new total amount.

$$\dfrac{\text{yellow}_{\text{new}}}{\text{total}_{\text{new}}} = \dfrac{14}{35} = \dfrac{2}{5} = 40\% \;\Rightarrow\; \textbf{(C)}$$

💡 Both $14$ and $35$ share the factor $7$, so $\tfrac{14}{35}$ simplifies to $\tfrac{2}{5}$, which every Grade 6 student knows as $40\%$.

[1] #7 6.RP.A.3 Convert the starting percent to liters of yellow. Of the $30$ liters of original

[2] #1 4.OA.A.3 Update the yellow amount after pouring in $5$ more liters of pure yellow tint. O

[3] #1 4.OA.A.3 Update the total volume. The added $5$ liters bumps the total mixture from $30$

[4] #7 6.RP.A.3 Convert back to a percent. Divide the new yellow amount by the new total amount.

Review

Reasonableness: Sanity-check the size of the answer. The yellow share started at $30\%$ and we added more yellow, so the new percent must be greater than $30\%$ — this rules out (A) $25$. But the addition is only $5$ liters compared to a total of $35$, so the jump cannot be huge — $50\%$ would mean half the mixture is yellow, which would need $17.5$ liters of yellow, far more than the $14$ we have. The actual jump from $30\%$ to $40\%$ is exactly $10$ percentage points, which fits the modest size of the pour. As a final check: the red tint and water keep their original amounts ($25\% \times 30 = 7.5$ and $45\% \times 30 = 13.5$), and $7.5 + 13.5 + 14 = 35$ matches the new total.

Alternative: Tool #1 (Draw a Diagram) on its own: draw a bar of length $30$ split into stripes of $7.5$ (red), $9$ (yellow), $13.5$ (water). Then glue a $5$-unit stripe onto the yellow end. The bar is now $35$ long with a yellow piece of $14$. Reading the yellow fraction off the bar gives $\tfrac{14}{35} = \tfrac{2}{5} = 40\%$ — answer (C).

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including percent problems (Converting $30\%$ of $30$ liters into $9$ liters of yellow tint, and converting the final ratio $\tfrac{14}{35}$ back into the percent $40\%$.)
4.OA.A.3 Solve multi-step word problems with whole numbers using the four operations (Adding $5$ liters to both the yellow amount ($9 + 5 = 14$) and the total amount ($30 + 5 = 35$) to update the mixture after the pour.)

⭐ When a mixture problem asks about percent, work in actual amounts first — turn $30\%$ into $9$ liters, do the addition, then turn the answer back into a percent. The numbers do the rest.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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