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

📏 Short solution 💡 2 insights

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Problem

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.

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.

How to solve

Strategy Solve an Easier Related Problem

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.

1STEP 1

Pool all the juice — pouring only moves it, so the total is conserved at $\frac{10}{3}$ glasses.

Total = 1 + 1 + 1 +

\frac{1}{3}

\frac{10}{3}

glasses

2STEP 2

Share that total equally among the four glasses; each should end at $\frac{5}{6}$ of a glass.

Target per glass = (

\frac{10}{3}

)/4 =

\frac{10}{3}

\frac{1}{4}

\frac{10}{12}

\frac{5}{6}

3STEP 3

Work backwards: each full glass must drop from 1 to $\frac{5}{6}$ , so pour out the difference, $\frac{1}{6}$ of a glass.

x = 1 -

\frac{5}{6}

\frac{6}{6}

\frac{5}{6}

\frac{1}{6}

4STEP 4

Check the fourth glass: three pours of $\frac{1}{6}$ add $\frac{1}{2}$ , lifting $\frac{1}{3}$ to $\frac{5}{6}$ — matching the other three. Answer (C).

\frac{1}{3}

+ 3 ·

\frac{1}{6}

\frac{2}{6}

\frac{3}{6}

\frac{5}{6}

→ (C)

Answer

frac{1}{6}

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}

÷ 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.

💡Key takeaway

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.

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