AMC 10 · 2019 · #1
Easy mode Grade 5Problem
Picture two containers. The first one is 65 full of water. The second one is empty.
Alicia pours all the water from the first container into the second container. After pouring, the second container is 43 full.
The same water filled 65 of the first container, and now fills 43 of the second container.
What is the ratio of the volume of the first container to the volume of the second container?
Pick an answer.
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.
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
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) $\tfrac{11}{12}$
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
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.
💡 Pick a round size so the fractions become whole numbers.
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 $9$.
💡 Same water, just measured in the first container's frame.
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.
💡 Divide by $\tfrac{5}{6}$ is the same as multiplying by $\tfrac{6}{5}$.
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).
💡 Simplify the ratio and match it to the answer choices.
5.NF.B.6 Pick a friendly number for $V_2$. Since denominators $6$ and $4$ have LCM $12$, 5.NF.B.6 All $9$ units of water came from the first container, where it was $\tfrac{5}{6} 5.NF.B.7 Solve for $V_1$ by undoing the $\tfrac{5}{6}$: multiply $9$ by $\tfrac{6}{5}$. S 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.1Add and subtract fractions with unlike denominators (Simplifying the ratio $\tfrac{54}{60}$ to $\tfrac{9}{10}$ by finding a common factor.)5.NF.B.6Solve 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.7Apply 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 $\tfrac{9}{10}$.
⭐ 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 $\tfrac{9}{10}$.
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