AMC 10 · 2023 · #1

Grade 5 rate-ratio

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

fraction-arithmeticmean-median-mode-rangeratio-proportion easier-related-problemwork-backwardsidentify-subproblems ↑ Prerequisites: fraction-arithmeticmean-median-mode-range

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Problem

Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?

Pick an answer.

(A)

$frac{1}{12}$

(B)

$frac{1}{4}$

(C)

$frac{1}{6}$

(D)

$frac{1}{8}$

(E)

$frac{2}{9}$

AMC 10 2023 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: Three glasses are full and a fourth holds only $\frac{1}{3}$ of a glass. Find the fraction $x$ that must be poured from each full glass into the fourth so that all four glasses end up with the same amount.

Givens: Glass 1, 2, 3 each hold $1$ glass of juice; Glass 4 holds $\frac{1}{3}$ of a glass; All four glasses are identical; Answer choices: (A) $\frac{1}{12}$, (B) $\frac{1}{4}$, (C) $\frac{1}{6}$, (D) $\frac{1}{8}$, (E) $\frac{2}{9}$

Unknowns: The fraction $x$ of a glass to pour from each of the first three into the fourth

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #11 Work Backwards

The easier problem hiding inside is just averaging: if all the juice were pooled and shared equally among $4$ identical glasses, how much would each glass get? Tool #9 reframes the whole pouring scene as "find the average level". Once we know the target level, Tool #11 (Work Backwards) finishes: each full glass must drop from $1$ to that target, and the amount poured out is the difference. A quick Tool #1 sketch of four bar-glasses makes the conservation visible without algebra.

Execute — Answer: C

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

Add up all the juice currently in the four glasses.
Pouring between glasses does not create or destroy juice, so the total stays the same before and after.

$$\text{Total} = 1 + 1 + 1 + \dfrac{1}{3} = \dfrac{10}{3} \text{ glasses}$$

💡 Adding $1$ three times and then $\frac{1}{3}$ is just adding fractions with the same denominator — a Grade 5 skill.

#9 Solve an Easier Related Problem 5.NF.B.7 Step 2

Split that total equally among the four glasses.
This is the target amount every glass should hold after the pouring is done.

$$\text{Target per glass} = \dfrac{10/3}{4} = \dfrac{10}{3} \cdot \dfrac{1}{4} = \dfrac{10}{12} = \dfrac{5}{6}$$

💡 Sharing a total equally among $4$ groups is just dividing a fraction by a whole number.

#11 Work Backwards 5.NF.A.1 Step 3

Each full glass starts at $1$ and must end at $\frac{5}{6}$.
Work backwards: the poured-out amount is what is removed to go from $1$ down to $\frac{5}{6}$.

$$x = 1 - \dfrac{5}{6} = \dfrac{6}{6} - \dfrac{5}{6} = \dfrac{1}{6}$$

💡 If you know the start ($1$) and the end ($\frac{5}{6}$), the missing piece is just the subtraction $1 - \frac{5}{6}$.

#1 Draw a Diagram 5.NF.A.2 Step 4

Verify by checking the fourth glass.
Three pourings of $\frac{1}{6}$ each go into glass $4$, so it gains $3 \cdot \frac{1}{6} = \frac{1}{2}$, ending at $\frac{1}{3} + \frac{1}{2} = \frac{5}{6}$ — same as the other three.
The answer is $\frac{1}{6}$, which is choice (C).

$$\dfrac{1}{3} + 3 \cdot \dfrac{1}{6} = \dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6} \Rightarrow \textbf{(C)}$$

💡 A picture of four equal bars at height $\frac{5}{6}$ confirms the pouring distributes correctly.

[1] #9 5.NF.A.1 Add up all the juice currently in the four glasses. Pouring between glasses does

[2] #9 5.NF.B.7 Split that total equally among the four glasses. This is the target amount every

[3] #11 5.NF.A.1 Each full glass starts at $1$ and must end at $\frac{5}{6}$. Work backwards: the

[4] #1 5.NF.A.2 Verify by checking the fourth glass. Three pourings of $\frac{1}{6}$ each go int

Review

Reasonableness: Sanity-check the magnitudes. The fourth glass is short by $\frac{5}{6} - \frac{1}{3} = \frac{1}{2}$ of a glass; splitting that shortage equally among the three donor glasses gives $\frac{1}{2} \div 3 = \frac{1}{6}$ per donor — matching the answer. Also, $\frac{1}{6}$ is small enough that each donor still has plenty left ($\frac{5}{6}$), which fits the physical picture.

Alternative: Tool #13 (Convert to Algebra). Let $x$ be the fraction poured from each full glass. The full glasses end at $1 - x$ and the short glass ends at $\frac{1}{3} + 3x$. Set them equal: $1 - x = \frac{1}{3} + 3x$, giving $\frac{2}{3} = 4x$, so $x = \frac{1}{6}$ — same answer with a heavier setup.

CCSS standards used (min grade 5)

5.NF.A.1 Add and subtract fractions with unlike denominators (Adding $1 + 1 + 1 + \frac{1}{3}$ to get the total $\frac{10}{3}$ and computing $1 - \frac{5}{6} = \frac{1}{6}$.)
5.NF.A.2 Solve word problems involving addition and subtraction of fractions (Verifying the fourth glass ends at $\frac{1}{3} + \frac{1}{2} = \frac{5}{6}$, matching the other three.)
5.NF.B.7 Apply and extend understanding of division to divide unit fractions by whole numbers (Dividing the total $\frac{10}{3}$ equally among $4$ glasses to find the target level $\frac{5}{6}$.)

⭐ This AMC 10 problem only needs Grade 5 fraction-sharing you already know — pool all the juice, divide by $4$, and pour out the leftover above that target.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 5)

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