2025 AMC 8 Problem 20

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To help you master 2025 AMC 8 Problem 20, this page includes:

The original AMC problem to try, plus the correct answer

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2025 AMC8 Problem 20

Solution

Solution 1

Relative Proportions Method

Instead of tracking the cheese turn-by-turn, compare the total amount each person eats. Because Dev always takes his turn immediately after Sarika, he always eats exactly half as much as she just did.

Ratios of Totals: If the amounts added to several running totals always maintain a fixed ratio at every step, then the final overall totals will also have that exact same fixed ratio.

Step 1: Define variables for the totals

Let the total fractions of the cheese eaten by Sarika, Dev, and Rajiv over the entire infinite process be represented by variables.

S = Total fraction eaten by Sarika

D = Total fraction eaten by Dev

R = Total fraction eaten by Rajiv

Step 2: Relate Dev’s total to Sarika’s total

Notice the sequence of turns. When Sarika takes her turn, she eats half of the available cheese, leaving the other half behind. Dev takes his turn next, eating half of that leftover amount. Because Dev eats half of what Sarika left, and Sarika left an amount exactly equal to what she ate, Dev always eats exactly half as much as Sarika did on the preceding turn. Because this rule applies to every single round, Dev’s final total must be half of Sarika’s total.

D = \frac{1}{2} S

Step 3: Relate Rajiv’s total to Dev’s total

By the exact same reasoning, Rajiv always takes his turn immediately after Dev. Rajiv eats half of the remaining cheese, meaning he always eats exactly half as much as Dev just did.

R = \frac{1}{2} D

R = \frac{1}{2} (\frac{1}{2} S) = \frac{1}{4} S

Step 4: Create an equation for the whole block of cheese

The problem states they stop when the cheese is ‘too small to see’. This implies the un-eaten remainder is microscopic and can be ignored. Mathematically, we assume the entire block is eaten. Set the sum of their individual total shares equal to 1 whole block.

S + D + R = 1

Step 5: Solve for Sarika’s fraction

Substitute Dev and Rajiv’s shares with their equivalents in terms of Sarika’s share, factor out S, and solve the equation.

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

S (1 + \frac{1}{2} + \frac{1}{4}) = 1

S (\frac{4}{4} + \frac{2}{4} + \frac{1}{4}) = 1

S (\frac{7}{4}) = 1

S = \frac{4}{7}

Correct answer: A. $\frac{4}{7}$
The final answer is exactly 4/7, corresponding to choice A. We can verify the total: if Sarika eats 4/7, Dev eats half of that (2/7), and Rajiv eats half of Dev’s (1/7). The sum is 4/7 + 2/7 + 1/7 = 7/7 = 1 whole block of cheese, confirming our logic is flawless.

Solution 2

Cycle-by-Cycle Grouping Method

Group the sequence of turns into repeated cycles of 3 (one turn for each person). Find what fraction of the cheese Sarika eats out of the total amount consumed in a single cycle.

Proportional Shares in Groups: If a total amount is divided into groups or stages, and you take the exact same fraction from every single group, you have taken that exact fraction of the overall total.

Step 1: Calculate the fractions eaten in the first cycle

Group the process into rounds of three turns. Find the exact fraction of the original block eaten by each person in the very first round.

Sarika eats \frac{1}{2}

Dev eats \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Rajiv eats \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

Step 2: Find the total cheese eaten in the first cycle

Add these three amounts together to find the total fraction of the cheese that was consumed during round 1.

Total eaten in round 1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}

Step 3: Find Sarika’s proportional share of the first round

Calculate what fraction of the consumed cheese in round 1 was eaten by Sarika. We divide her amount by the total amount eaten in that round.

Sarika’s share of round 1 = \frac{\frac{1}{2}}{\frac{7}{8}}

= \frac{1}{2} \times \frac{8}{7} = \frac{4}{7}

Step 4: Generalize to all subsequent rounds

In the second round, the three people start with the remaining one-eighth of the cheese. However, they will again follow the same rules, taking one half, one quarter, and one eighth of that remaining amount. Because the fractions shrink proportionally, Sarika will again eat exactly 4/7 of the cheese consumed in that round.

Cheese consumed in round 2 = \frac{1}{16} + \frac{1}{32} + \frac{1}{64} = \frac{7}{64}

Sarika’s share of round 2 = \frac{\frac{1}{16}}{\frac{7}{64}} = \frac{4}{7}

Step 5: Conclude the overall fraction

Since Sarika eats exactly 4/7 of the cheese consumed in every single round, and eventually the entire block of cheese is consumed, her total share of the entire block must also be 4/7.

Overall share = \frac{4}{7}

Correct answer: A. $\frac{4}{7}$
The final answer is 4/7, which is choice A. Checking the complementary proportion: the non-Sarika fraction of any round is 3/7 (since Dev and Rajiv eat 2/7 and 1/7 of the round respectively). Sarika’s 4/7 plus the others’ 3/7 perfectly accounts for all the eaten cheese.

Solution 3

Infinite Geometric Series Method

Track the exact fraction of the original cheese block that Sarika eats on each of her individual turns and add them all together using an infinite sequence.

Sum of an Infinite Geometric Series: If a sequence starts with a first term $a$ and each subsequent term is multiplied by a common ratio $r$ (where $- 1 < r < 1$ ), the sum of all infinitely many terms is $S = \frac{a}{1 - r}$ .

Step 1: Identify Sarika’s first few portions

Calculate the exact fraction of the original block that Sarika eats on her first few turns. On her first turn, she eats half. After Dev and Rajiv take their turns, the remaining cheese is one-eighth of the original. On her second turn, Sarika eats half of that remaining one-eighth.

a_{1} = \frac{1}{2}

a_{2} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}

Step 2: Determine the common ratio

Notice that Sarika’s portion on any turn is exactly one-eighth of her portion on her previous turn. This makes sense because a full cycle of three turns leaves exactly one-eighth of the cheese that was available at the start of the cycle.

r = \frac{1}{8}

Step 3: Set up the infinite geometric series

Because they stop when the cheese is ‘too small to see’, we model this as an infinite sequence. Sarika’s total share is the sum of this infinite geometric series.

S = \frac{1}{2} + \frac{1}{16} + \frac{1}{128} + \dots

Step 4: Apply the sum formula

Use the formula for the sum of an infinite geometric series with our identified first term and common ratio.

S = \frac{a}{1 - r}

S = \frac{\frac{1}{2}}{1 - \frac{1}{8}}

S = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{4}{7}

Correct answer: A. $\frac{4}{7}$
The final answer is 4/7, corresponding to choice A. We can sum the series for Dev (first term 1/4, ratio 1/8) to get 2/7, and for Rajiv (first term 1/8, ratio 1/8) to get 1/7. Summing their independent infinite series yields 4/7 + 2/7 + 1/7 = 1, verifying the whole block is properly accounted for.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Ratios of Totals in Repeating Processes:

If multiple participants take shares in a repeating sequence, and the ratio between their individual shares remains constant at every step, then the ratio of their final overall totals will be exactly the same.

Applies when the rules governing the process ensure a fixed proportion between the amounts claimed by different participants on consecutive turns. Dev always eats half of the exact amount Sarika just left behind, which means he always eats exactly half of what Sarika just ate. This fixed 2-to-1 ratio applies to every single turn they take.

Proportional Shares in Groups:

If a total amount is divided into stages or cycles, and you claim the exact same fraction of the consumed amount in every cycle, your share of the total consumed amount will be that exact fraction.

Requires the sequence to be naturally divisible into cycles where the internal proportions remain constant. In every round of 3 turns, Sarika eats 4/7 of the cheese consumed in that specific round. Therefore, she eats 4/7 of the entire cheese block overall.

Sum of an Infinite Geometric Series:

S = \frac{a}{1 - r}

S is the total sum, a is the first term, and r is the common ratio (the constant multiplier between consecutive terms, where -1 < r < 1). Sarika's portions form an infinite sequence (1/2, 1/16, 1/128...) where each portion is exactly 1/8 of her previous portion.

Key Takeaway: Remember This Strategy for Next Time

A repeating process where a constant fraction of the remainder is removed at each step by different participants.

Try: Look for a fixed algebraic ratio between the final totals of the participants instead of calculating each individual step.

Why: If the rules dictate that one person always takes exactly half as much as another person, their final totals will also follow that exact 2-to-1 ratio, allowing you to quickly solve using a simple equation like S + S/2 + S/4 = 1.

A sequence of actions that repeats in identical rounds or cycles.

Try: Analyze a single complete cycle and calculate what proportion of that cycle belongs to the target person.

Why: If a person claims the exact same fraction of the total in every single cycle, their overall share of the final sum will be exactly that fraction, allowing you to ignore the infinite nature of the problem entirely.

A mathematical process that continues indefinitely and shrinks by a constant factor.

Try: Use the infinite geometric series formula $S = \frac{a}{1 - r}$ .

Why: Processes that theoretically 'never end' (or stop when a quantity is 'too small to see') but get proportionally smaller will converge to a specific, finite sum.

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