AMC 10 · 2023 · #1

Grade 5 rate-ratio
fraction-arithmeticmean-median-mode-rangeratio-proportion easier-related-problemwork-backwardsidentify-subproblems ↑ Prerequisites: fraction-arithmeticmean-median-mode-range
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Problem
Three glasses are full and a fourth holds only one third of a glass. The same fraction must be poured from each full glass into the fourth so that all four end up with the same amount. Find that fraction.

Pick an answer.

(A)
$\frac{1}{12}$
(B)
$\frac{1}{4}$
(C)
$\frac{1}{6}$
(D)
$\frac{1}{8}$
(E)
$\frac{2}{9}$
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

Find the total

The total never changes.

Total = 1 + 1 + 1 + 1/3 = 10/3 glasses
2STEP 2

Find the target per glass

Dividing by four gives the target.

Target per glass = 10/3/4 = 10/3 · 1/4 = 10/12 = 5/6
3STEP 3

Find how much leaves

What leaves a full glass is the answer.

x = 1 - 5/6 = 6/6 - 5/6 = 1/6
4STEP 4

Check it

Checking confirms one sixth.

1/3 + 3 · 1/6 = 2/6 + 3/6 = 5/6 → (C)
Answer
1/6
Sanity-check the magnitudes. The fourth glass is short by 5/6 - 1/3 = 1/2 of a glass; splitting that shortage equally among the three donor glasses gives 1/2 ÷ 3 = 1/6 per donor — matching the answer. Also, 1/6 is small enough that each donor still has plenty left (5/6), which fits the physical picture.
💡Key takeaway

This AMC 12 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.