AMC 8 · 2025 · #20
Grade 6 rate-ratioalgebraProblem
Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The cheese starts as one whole block (call it $1$); Each person eats exactly half of whatever is left when their turn begins; Turn order: Sarika $\to$ Dev $\to$ Rajiv $\to$ Sarika $\to \dots$; The process keeps repeating until essentially nothing is left; Answer choices: (A) $\tfrac{4}{7}$, (B) $\tfrac{3}{5}$, (C) $\tfrac{2}{3}$, (D) $\tfrac{3}{4}$, (E) $\tfrac{7}{8}$
Unknowns: The total fraction of the original cheese that Sarika ends up eating
Understand
Restated: 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.
Givens: The cheese starts as one whole block (call it $1$); Each person eats exactly half of whatever is left when their turn begins; Turn order: Sarika $\to$ Dev $\to$ Rajiv $\to$ Sarika $\to \dots$; The process keeps repeating until essentially nothing is left; Answer choices: (A) $\tfrac{4}{7}$, (B) $\tfrac{3}{5}$, (C) $\tfrac{2}{3}$, (D) $\tfrac{3}{4}$, (E) $\tfrac{7}{8}$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #16 Change Focus / Complement
Direct summation of an infinite series $\tfrac{1}{2} + \tfrac{1}{16} + \tfrac{1}{128} + \dots$ 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.
Execute — Answer: A
5.NF.B.4 Step 1 - Play out one full cycle on a block of size $1$ to see exactly what each person eats.
- Sarika eats $\tfrac{1}{2}$, leaving $\tfrac{1}{2}$.
- Dev eats half of that, $\tfrac{1}{4}$, leaving $\tfrac{1}{4}$.
- Rajiv eats half of that, $\tfrac{1}{8}$, leaving $\tfrac{1}{8}$.
💡 Working out one small cycle by hand turns an abstract "forever" problem into something a $5$th grader can compute with fraction multiplication.
6.RP.A.1 Step 2 - Compare the three bites of that cycle.
- Multiplying each by $8$ to clear the fractions, Sarika ate $4$ parts, Dev ate $2$ parts, Rajiv ate $1$ part, and $1$ part was left for the next cycle.
- So inside each cycle Sarika $:$ Dev $:$ Rajiv $= 4 : 2 : 1$.
💡 The pattern is just three consecutive halvings, so each person gets exactly half of what the previous person got — a classic Grade 6 ratio.
6.RP.A.1 Step 3 - Notice that this $4 : 2 : 1$ ratio is the SAME for every cycle.
- After cycle $1$ only $\tfrac{1}{8}$ of the original is left, so cycle $2$ is just a scaled-down copy: Sarika gets $\tfrac{4}{7}$ of $\tfrac{1}{8}$, Dev gets $\tfrac{2}{7}$ of $\tfrac{1}{8}$, Rajiv gets $\tfrac{1}{7}$ of $\tfrac{1}{8}$ — same ratio.
- The ratio of total cheese eaten by the three of them over the whole game is therefore $4 : 2 : 1$.
💡 If every cycle splits its share in the same ratio, the total over all cycles splits in that ratio too.
6.RP.A.3 Step 4 - Change focus from "sum Sarika's infinite bites" to "split the whole block by the $4 : 2 : 1$ ratio." Because the leftover shrinks to $0$, the three of them together eat the entire block of size $1$.
- Total parts $= 4 + 2 + 1 = 7$, so each part is $\tfrac{1}{7}$ of the block.
💡 Splitting a whole into a known ratio is exactly the Grade 6 "share in the ratio" move — no infinite series needed.
6.RP.A.3 Step 5 - Sarika owns $4$ of the $7$ equal parts, so she eats $\tfrac{4}{7}$ of the original block.
- This matches choice (A).
💡 Her share is her ratio piece divided by the sum of all ratio pieces — basic Grade 6 ratio arithmetic.
5.NF.B.4 Play out one full cycle on a block of size $1$ to see exactly what each person e 6.RP.A.1 Compare the three bites of that cycle. Multiplying each by $8$ to clear the frac 6.RP.A.1 Notice that this $4 : 2 : 1$ ratio is the SAME for every cycle. After cycle $1$ 6.RP.A.3 Change focus from "sum Sarika's infinite bites" to "split the whole block by the 6.RP.A.3 Sarika owns $4$ of the $7$ equal parts, so she eats $\tfrac{4}{7}$ of the origin Review
Reasonableness: Sarika goes first and always takes half of what is currently there, so she should eat the biggest share — more than $\tfrac{1}{3}$ but less than $\tfrac{2}{3}$ (since the other two together also get something on every cycle). $\tfrac{4}{7} \approx 0.571$ sits comfortably in that range, while choices (C) $\tfrac{2}{3} \approx 0.667$ and (D) $\tfrac{3}{4}$ would crowd Dev and Rajiv out. A sanity check on partial sums: $\tfrac{1}{2} + \tfrac{1}{16} = \tfrac{9}{16} = 0.5625$, already very close to $\tfrac{4}{7} \approx 0.5714$, and the next term $\tfrac{1}{128}$ pushes it to $\tfrac{73}{128} \approx 0.5703$ — converging to $\tfrac{4}{7}$ exactly.
Alternative: Tool #6 (Guess and Check) on the choices: by symmetry of the cycle, Dev gets half of Sarika's share and Rajiv gets a quarter of Sarika's share, so Sarika $+$ Dev $+$ Rajiv $= S(1 + \tfrac{1}{2} + \tfrac{1}{4}) = \tfrac{7}{4} S = 1$, giving $S = \tfrac{4}{7}$. Same answer (A) from a different angle, confirming the ratio reasoning.
CCSS standards used (min grade 6)
5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Computing each person's first-cycle share by halving: $\tfrac{1}{2}, \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}, \tfrac{1}{2} \times \tfrac{1}{4} = \tfrac{1}{8}$.)6.RP.A.1Understand the concept of a ratio and use ratio language (Expressing the three first-cycle bites as the ratio $\tfrac{1}{2} : \tfrac{1}{4} : \tfrac{1}{8} = 4 : 2 : 1$ and recognizing this ratio is preserved in every later cycle.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Sharing the whole block in the ratio $4 : 2 : 1$ to get Sarika's portion $\tfrac{4}{4+2+1} = \tfrac{4}{7}$ without summing an infinite series.)
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — sharing a whole in a $4 : 2 : 1$ ratio — you already know!
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — sharing a whole in a $4 : 2 : 1$ ratio — you already know!