AMC 8 · 2004 · #16

Grade 5 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

fraction-arithmeticfraction-multiplicationratio-proportion identify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 2 insights

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Problem

Two 600 mL pitchers hold orange juice. One is

\frac{1}{3}

full of juice and the other is

\frac{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?

Pick an answer.

(A)

$\frac18$

(B)

$\frac{3}{16}$

(C)

$\frac{11}{30}$

(D)

$\frac{11}{19}$

(E)

$\frac{11}{15}$

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Break into Subproblems

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.

1STEP 1

First pitcher: $\frac{1}{3}$ of 600 mL is 200 mL of juice.

OJ₁ =

\frac{1}{3}

× 600 = 200 mL

2STEP 2

Second pitcher: $\frac{2}{5}$ of 600 mL is 240 mL of juice.

OJ₂ =

\frac{2}{5}

× 600 =

\frac{1200}{5}

= 240 mL

3STEP 3

Add both pitchers' juice: 200 + 240 = 440 mL in the container.

OJ_total = 200 + 240 = 440 mL

4STEP 4

Each pitcher is topped off, so the container holds 600 + 600 = 1200 mL of liquid.

V_total = 600 + 600 = 1200 mL

5STEP 5

Juice over total: $\frac{440}{1200}$ reduces to $\frac{11}{30}$ → (C).

OJ_total/V_total =

\frac{440}{1200}

\frac{44}{120}

\frac{11}{30}

→ (C)

Answer

\frac{11}{30}

Bounds check: each pitcher alone is less than half juice (

\frac{1}{3}

and

\frac{2}{5}

are both under

\frac{1}{2}

), so the combined mixture must also be under

\frac{1}{2}

. Our answer

\frac{11}{30}

≈ 0.367 sits between

\frac{1}{3}

and

\frac{2}{5}

, exactly where the average of two equal-size pitchers should land. Choices (D)

\frac{11}{19}

≈ 0.58 and (E)

\frac{11}{15}

≈ 0.73 are both above

\frac{1}{2}

, so they are immediately out.

💡Key takeaway

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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