AMC 10 · 2019 · #1

Grade 5 geometry-3d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

ratio-proportionfraction-multiplicationfraction-arithmetic identify-subproblems ↑ Prerequisites: fraction-arithmeticratio-proportion

📏 Short solution 💡 2 insights

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Problem

Alicia had two containers. The first was $\tfrac{5}{6}$ full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\tfrac{3}{4}$ full of water. What is the ratio of the volume of the first container to the volume of the second container?

Pick an answer.

(A)

$\frac{5}{8}$

(B)

$\frac{4}{5}$

(C)

$\frac{7}{8}$

(D)

$\frac{9}{10}$

(E)

$\frac{11}{12}$

AMC 10 2019 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: Alicia has two containers. The first starts $\tfrac{5}{6}$ full of water; the second is empty. She pours all the water from the first into the second, which then becomes $\tfrac{3}{4}$ full. Find the ratio of the first container's volume to the second container's volume.

Givens: Water in the first container fills $\tfrac{5}{6}$ of its volume; After pouring, the same water fills $\tfrac{3}{4}$ of the second container's volume; Answer choices: (A) $\tfrac{5}{8}$, (B) $\tfrac{4}{5}$, (C) $\tfrac{7}{8}$, (D) $\tfrac{9}{10}$, (E) $\tfrac{11}{12}$

Unknowns: The ratio $\dfrac{V_1}{V_2}$ of the first container's volume to the second container's

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #8 Analyze the Units, #3 Eliminate Possibilities

Tool #9 (Easier Problem): pick a friendly number for the second container's volume (say $V_2 = 12$, the LCM of $6$ and $4$) so both fractions give whole-number amounts. Then read $V_1$ off directly. Tool #8 (Units): water amount poured equals water amount received, so $V_1 \cdot \tfrac{5}{6} = V_2 \cdot \tfrac{3}{4}$ — solving for $V_1 / V_2$ falls right out. Tool #3 (Eliminate): the first container holds less water at $\tfrac{5}{6}$ full than the second at $\tfrac{3}{4}$ full only if $V_1 < V_2$, so the ratio is less than $1$ — all five choices satisfy this, but the clean fraction match picks (D).

Execute — Answer: D

#9 Solve an Easier Related Problem 5.NF.B.6 Step 1

Pick a friendly number for $V_2$.
Since denominators $6$ and $4$ have LCM $12$, let $V_2 = 12$ units.
Then $\tfrac{3}{4}$ of $12$ is $9$ units of water in the second container after pouring.

$$V_2 = 12 \;\Rightarrow\; \text{water poured in} = \tfrac{3}{4} \cdot 12 = 9$$

💡 Pick a round size so the fractions become whole numbers.

#8 Analyze the Units 5.NF.B.6 Step 2

All $9$ units of water came from the first container, where it was $\tfrac{5}{6}$ full.
So $\tfrac{5}{6}$ of $V_1$ equals $9$.

$$\tfrac{5}{6} \cdot V_1 = 9$$

💡 Same water, just measured in the first container's frame.

#9 Solve an Easier Related Problem 5.NF.B.7 Step 3

Solve for $V_1$ by undoing the $\tfrac{5}{6}$: multiply $9$ by $\tfrac{6}{5}$.
So $V_1 = 9 \cdot \tfrac{6}{5} = \tfrac{54}{5} = 10.8$ units.

$$V_1 = 9 \cdot \tfrac{6}{5} = \tfrac{54}{5}$$

💡 Divide by $\tfrac{5}{6}$ is the same as multiplying by $\tfrac{6}{5}$.

#3 Eliminate Possibilities 5.NF.A.1 Step 4

Now form the ratio $V_1 : V_2 = \tfrac{54}{5} : 12$.
Multiply both sides by $5$ to clear the fraction: $54 : 60$, then simplify by dividing both by $6$ to get $9 : 10$.
So the ratio is $\tfrac{9}{10}$, matching choice (D).

$$\dfrac{V_1}{V_2} = \dfrac{54/5}{12} = \dfrac{54}{60} = \dfrac{9}{10} \;\Rightarrow\; \textbf{(D)}$$

💡 Simplify the ratio and match it to the answer choices.

[1] #9 5.NF.B.6 Pick a friendly number for $V_2$. Since denominators $6$ and $4$ have LCM $12$,

[2] #8 5.NF.B.6 All $9$ units of water came from the first container, where it was $\tfrac{5}{6}

[3] #9 5.NF.B.7 Solve for $V_1$ by undoing the $\tfrac{5}{6}$: multiply $9$ by $\tfrac{6}{5}$. S

[4] #3 5.NF.A.1 Now form the ratio $V_1 : V_2 = \tfrac{54}{5} : 12$. Multiply both sides by $5$

Review

Reasonableness: Sanity check: the first container at $\tfrac{5}{6}$ full holds the same water as the second at $\tfrac{3}{4}$ full. Since $\tfrac{5}{6} > \tfrac{3}{4}$, the same amount of water fills more of the first container — so the first must be smaller. The ratio $\tfrac{9}{10} < 1$ confirms this. Plugging back: $V_1 = 10.8$, $\tfrac{5}{6} \cdot 10.8 = 9$; $V_2 = 12$, $\tfrac{3}{4} \cdot 12 = 9$. Both sides match.

Alternative: Tool #13 (Algebra): set $\tfrac{5}{6} V_1 = \tfrac{3}{4} V_2$ directly, then $\dfrac{V_1}{V_2} = \dfrac{3/4}{5/6} = \dfrac{3}{4} \cdot \dfrac{6}{5} = \dfrac{18}{20} = \dfrac{9}{10}$. Same answer with no number-picking.

CCSS standards used (min grade 5)

5.NF.A.1 Add and subtract fractions with unlike denominators (Simplifying the ratio $\tfrac{54}{60}$ to $\tfrac{9}{10}$ by finding a common factor.)
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers (Computing $\tfrac{3}{4} \cdot 12 = 9$ units of water in the second container.)
5.NF.B.7 Apply and extend understanding of division to divide unit fractions by whole numbers (Solving $\tfrac{5}{6} \cdot V_1 = 9$ by dividing $9$ by $\tfrac{5}{6}$ (i.e., multiplying by $\tfrac{6}{5}$).)

⭐ This AMC 10 problem only needs Grade 5 fraction multiplication and division you already know — pick the second jar to be $12$ units, then the water is $9$ units, the first jar is $10.8$ units, and the ratio is $\tfrac{9}{10}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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