AMC 8 · 2004 · #16
Easy mode Grade 5Problem
Imagine two pitchers. Each pitcher can hold 600 mL when full.
The first pitcher has orange juice in it. The juice fills 31 of the pitcher.
The second pitcher also has orange juice. This time the juice fills 52 of the pitcher.
Now water is poured into each pitcher until both pitchers are completely full. Then both pitchers are emptied into one big container.
In the big container, what fraction of the mixture is orange juice?
Pick an answer.
AMC 8 2004 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: Two $600$ mL pitchers hold orange juice. One is $\tfrac{1}{3}$ full of juice and the other is $\tfrac{2}{5}$ full. Each pitcher is then topped off with water, and both pitchers are emptied into one large container. What fraction of the final mixture is orange juice?
Givens: Each pitcher has capacity $600$ mL; Pitcher 1 is $\tfrac{1}{3}$ full of orange juice; Pitcher 2 is $\tfrac{2}{5}$ full of orange juice; Water is added to fill each pitcher completely, then both are combined; Answer choices: (A) $\tfrac{1}{8}$, (B) $\tfrac{3}{16}$, (C) $\tfrac{11}{30}$, (D) $\tfrac{11}{19}$, (E) $\tfrac{11}{15}$
Unknowns: The fraction of the combined mixture that is orange juice
Understand
Restated: Two $600$ mL pitchers hold orange juice. One is $\tfrac{1}{3}$ full of juice and the other is $\tfrac{2}{5}$ full. Each pitcher is then topped off with water, and both pitchers are emptied into one large container. What fraction of the final mixture is orange juice?
Givens: Each pitcher has capacity $600$ mL; Pitcher 1 is $\tfrac{1}{3}$ full of orange juice; Pitcher 2 is $\tfrac{2}{5}$ full of orange juice; Water is added to fill each pitcher completely, then both are combined; Answer choices: (A) $\tfrac{1}{8}$, (B) $\tfrac{3}{16}$, (C) $\tfrac{11}{30}$, (D) $\tfrac{11}{19}$, (E) $\tfrac{11}{15}$
Plan
Primary tool: #7 Break into Subproblems
Secondary: #4 Introduce a Variable
The question asks for one fraction, but to find it we need two separate quantities: how much orange juice is in the container, and how much liquid is in the container in total. Tool #7 (Break into Subproblems) splits the work into those two parallel computations. Tool #4 (Introduce a Variable) is a light support — we treat each pitcher's capacity as the same $600$ mL so the two juice amounts can be added directly. Once the two subtotals are in hand, the final fraction is one division and a reduction.
Execute — Answer: C
5.NF.B.4 Step 1 - Subproblem 1: orange juice in the first pitcher.
- A fraction of a whole means multiplication.
- The first pitcher is $\tfrac{1}{3}$ full of juice, so it holds $\tfrac{1}{3}$ of $600$ mL.
💡 Grade 5 "fraction of a whole" — split $600$ into $3$ equal parts and take one part.
5.NF.B.4 Step 2 - Subproblem 2: orange juice in the second pitcher.
- Same idea, with $\tfrac{2}{5}$ of $600$ mL this time.
💡 Split $600$ into $5$ equal parts of $120$ mL each, then take two of them.
4.NBT.B.4 Step 3 - Subproblem 3: total juice in the container.
- Adding both pitchers' juice into one container just sums the two amounts.
💡 Combining contents = whole-number addition.
4.NBT.B.4 Step 4 - Subproblem 4: total mixture volume.
- Each pitcher is topped off with water, so each one contributes its full $600$ mL to the container.
💡 The water fills whatever space the juice did not, so each pitcher gives a full $600$ mL.
5.NF.B.3 Step 5 - Combine the subproblems.
- The juice fraction of the mixture is the juice volume divided by the total volume.
- Reduce by dividing out common factors.
💡 Grade 5 "fraction as division": $440$ parts juice out of $1200$ parts mixture, reduced to lowest terms by dividing by $10$ then by $4$.
5.NF.B.4 Subproblem 1: orange juice in the first pitcher. A fraction of a whole means mul 5.NF.B.4 Subproblem 2: orange juice in the second pitcher. Same idea, with $\tfrac{2}{5}$ 4.NBT.B.4 Subproblem 3: total juice in the container. Adding both pitchers' juice into one 4.NBT.B.4 Subproblem 4: total mixture volume. Each pitcher is topped off with water, so ea 5.NF.B.3 Combine the subproblems. The juice fraction of the mixture is the juice volume d Review
Reasonableness: Bounds check: each pitcher alone is less than half juice ($\tfrac{1}{3}$ and $\tfrac{2}{5}$ are both under $\tfrac{1}{2}$), so the combined mixture must also be under $\tfrac{1}{2}$. Our answer $\tfrac{11}{30} \approx 0.367$ sits between $\tfrac{1}{3}$ and $\tfrac{2}{5}$, exactly where the average of two equal-size pitchers should land. Choices (D) $\tfrac{11}{19} \approx 0.58$ and (E) $\tfrac{11}{15} \approx 0.73$ are both above $\tfrac{1}{2}$, so they are immediately out.
Alternative: Tool #16 (Transform): because the two pitchers are equal in size, the juice fraction of the mixture is just the average of the two juice fractions. Compute $\tfrac{1}{2}\left(\tfrac{1}{3} + \tfrac{2}{5}\right) = \tfrac{1}{2} \cdot \tfrac{5 + 6}{15} = \tfrac{1}{2} \cdot \tfrac{11}{15} = \tfrac{11}{30}$. Same answer with no mL units.
CCSS standards used (min grade 5)
5.NF.B.4Apply and extend previous understandings of multiplication to multiply a fraction by a whole number (Computing each pitcher's juice volume as a fraction of $600$ mL: $\tfrac{1}{3} \times 600 = 200$ and $\tfrac{2}{5} \times 600 = 240$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Adding the two juice volumes ($200 + 240 = 440$) and the two pitcher volumes ($600 + 600 = 1200$).)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Forming the final fraction $\tfrac{440}{1200}$ and reducing to $\tfrac{11}{30}$.)
⭐ Solve two small questions first — how much juice, how much liquid in all — then put one over the other and simplify.
⭐ Solve two small questions first — how much juice, how much liquid in all — then put one over the other and simplify.
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