AMC 8 · 2025 · #20

Grade 6 rate-ratioalgebra

Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share

sequences-geometricratio-proportionfraction-multiplicationpattern-recognition easier-related-problempattern-recognition ↑ Prerequisites: fraction-multiplicationratio-proportion

📏 Medium solution 💡 3 insights

Problem

Sarika, Dev, and Rajiv take turns (in that order, repeating) eating half of the cheese that is currently left. Sarika goes first, then Dev, then Rajiv, then back to Sarika, and so on forever (they stop only when the cheese is too small to see). Find the total fraction of the original block that Sarika eats across all of her turns.

Pick an answer.

(A)

$\frac{4}{7}$

(B)

$\frac{3}{5}$

(C)

$\frac{2}{3}$

(D)

$\frac{3}{4}$

(E)

$\frac{7}{8}$

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Direct summation of an infinite series $\frac{1}{2}$ + $\frac{1}{16}$ + $\frac{1}{128}$ + … is high-school material. Tool #9 (Easier Problem) says: just work out one full Sarika-Dev-Rajiv cycle on a whole block. Tool #5 (Pattern) then notices that inside any cycle the three of them eat the cheese in the fixed ratio 4 : 2 : 1, no matter how big the cheese-at-cycle-start is. Tool #16 (Change Focus) reframes the question: instead of summing infinitely many of Sarika's bites, use the fact that the three of them together eat the whole block, and split the whole block by that same fixed ratio. This avoids the infinite series entirely and keeps the math at elementary-school ratio reasoning.

1STEP 1

Play out one full round on a size-1 block: Sarika eats $\frac{1}{2}$ , Dev eats $\frac{1}{4}$ , Rajiv eats $\frac{1}{8}$ , leaving $\frac{1}{8}$ .

Sarika =

\frac{1}{2}

, Dev =

\frac{1}{4}

, Rajiv =

\frac{1}{8}

→ remaining =

\frac{1}{8}

2STEP 2

Scale those three bites by 8: Sarika 4 parts, Dev 2, Rajiv 1, with 1 part left. So each round splits 4 : 2 : 1.

\frac{1}{2}

\frac{1}{4}

\frac{1}{8}

= 4 : 2 : 1

3STEP 3

That 4 : 2 : 1 split repeats every round, since each later round is just a shrunk copy. So the whole game's totals share the same ratio.

cycle k: Sarika:Dev:Rajiv = 4:2:1 (always)

4STEP 4

Reframe: instead of summing bites, split the whole block by 4 : 2 : 1. Parts total 4 + 2 + 1 = 7, so one part is $\frac{1}{7}$ of the block.

4 + 2 + 1 = 7 → 1 part =

\frac{1}{7}

of the block

5STEP 5

Sarika owns 4 of the 7 equal parts, so she eats $\frac{4}{7}$ of the block — choice (A).

Sarika's total =

\frac{4}{4+2+1}

\frac{4}{7}

→ (A)

Answer

\frac{4}{7}

Sarika goes first and always takes half of what is currently there, so she should eat the biggest share — more than

\frac{1}{3}

but less than

\frac{2}{3}

(since the other two together also get something on every cycle).

\frac{4}{7}

≈ 0.571 sits comfortably in that range, while choices (C)

\frac{2}{3}

≈ 0.667 and (D)

\frac{3}{4}

would crowd Dev and Rajiv out. A sanity check on partial sums:

\frac{1}{2}

\frac{1}{16}

\frac{9}{16}

= 0.5625, already very close to

\frac{4}{7}

≈ 0.5714, and the next term

\frac{1}{128}

pushes it to

\frac{73}{128}

≈ 0.5703 — converging to

\frac{4}{7}

exactly.

💡Key takeaway

This AMC 8 problem only needs Grade 6 ratio reasoning — sharing a whole in a 4 : 2 : 1 ratio — you already know!

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