AMC 8 · 2019 · #8
Grade 6 arithmeticProblem
Gilda has a bag of marbles. She gives 20% of them to her friend Pedro. Then Gilda gives 10% of what is left to another friend, Ebony. Finally, Gilda gives 25% of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?
Pick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Gilda starts with a full bag of marbles. She gives $20\%$ of the bag to Pedro, then $10\%$ of what is left to Ebony, then $25\%$ of what is still left to Jimmy. We want the percentage of the original bag that Gilda still holds for herself after all three gifts.
Givens: Gilda begins with $100\%$ of her marbles; Step 1: she gives away $20\%$ of the current bag (to Pedro); Step 2: she gives away $10\%$ of what remains (to Ebony); Step 3: she gives away $25\%$ of what remains (to Jimmy); Answer choices: (A) $20$, (B) $33\tfrac{1}{3}$, (C) $38$, (D) $45$, (E) $54$
Unknowns: The percentage of Gilda's original bag that she still has after the three gifts
Understand
Restated: Gilda starts with a full bag of marbles. She gives $20\%$ of the bag to Pedro, then $10\%$ of what is left to Ebony, then $25\%$ of what is still left to Jimmy. We want the percentage of the original bag that Gilda still holds for herself after all three gifts.
Givens: Gilda begins with $100\%$ of her marbles; Step 1: she gives away $20\%$ of the current bag (to Pedro); Step 2: she gives away $10\%$ of what remains (to Ebony); Step 3: she gives away $25\%$ of what remains (to Jimmy); Answer choices: (A) $20$, (B) $33\tfrac{1}{3}$, (C) $38$, (D) $45$, (E) $54$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The trap in this problem is that each percentage applies to a *different* total (the current bag, not the original). Tool #9 (Easier Related Problem) makes this concrete by pretending Gilda starts with exactly $100$ marbles — then every percentage becomes a plain whole number we can track. Tool #7 (Identify Subproblems) splits the journey into three clean one-step "give and keep" calculations, one per friend. Tool #3 (Eliminate Possibilities) is a nice safety net at the end: among (A)$20$, (B)$33\tfrac{1}{3}$, (C)$38$, (D)$45$, (E)$54$, only one matches the value we compute, so we can confirm the choice directly.
Execute — Answer: E
6.RP.A.3 Step 1 - Pretend Gilda starts with exactly $100$ marbles.
- This isn't given, but the answer is a *percentage of her own bag*, so the bag's size doesn't change the answer — and using $100$ turns every percentage into an easy whole number.
💡 Setting the whole equal to $100$ is the standard Grade 6 trick for percent problems: "percent" literally means "per $100$."
6.RP.A.3 Step 2 - Stage 1 — Pedro.
- Gilda gives $20\%$ of $100 = 20$ marbles to Pedro, so she keeps $100 - 20 = 80$ marbles.
- (Equivalently, she keeps $80\%$ of what she had.)
💡 Taking $20\%$ of $100$ and subtracting is one clean subproblem — exactly the Tool #7 move.
6.RP.A.3 Step 3 - Stage 2 — Ebony.
- The new "whole" is $80$ marbles, not the original $100$.
- Ebony gets $10\%$ of $80 = 8$ marbles, so Gilda keeps $80 - 8 = 72$ marbles.
💡 The key Grade 6 idea: each percent is applied to the *current* bag, so the base changes between stages.
5.NF.B.4 Step 4 - Stage 3 — Jimmy.
- The whole is now $72$ marbles.
- Jimmy gets $25\%$ of $72$.
- Use the fact that $25\% = \tfrac{1}{4}$ to avoid decimals: $\tfrac{1}{4} \times 72 = 18$.
- Gilda keeps $72 - 18 = 54$ marbles.
💡 Recognizing $25\% = \tfrac{1}{4}$ and multiplying a whole number by a unit fraction is a Grade 5 fraction-times-whole-number move.
6.RP.A.3 Step 5 - Since we started with $100$ marbles and Gilda kept $54$, she has $\tfrac{54}{100} = 54\%$ of the original bag.
- Scanning the choices, only (E) $54$ matches, so this is the answer.
💡 Once the kept amount lines up with exactly one choice, Tool #3 (Eliminate) confirms the pick with no extra work.
6.RP.A.3 Pretend Gilda starts with exactly $100$ marbles. This isn't given, but the answe 6.RP.A.3 Stage 1 — Pedro. Gilda gives $20\%$ of $100 = 20$ marbles to Pedro, so she keeps 6.RP.A.3 Stage 2 — Ebony. The new "whole" is $80$ marbles, not the original $100$. Ebony 5.NF.B.4 Stage 3 — Jimmy. The whole is now $72$ marbles. Jimmy gets $25\%$ of $72$. Use t 6.RP.A.3 Since we started with $100$ marbles and Gilda kept $54$, she has $\tfrac{54}{100 Review
Reasonableness: A quick sanity check: Gilda gave away three chunks, but each chunk was at most a quarter of the current bag, so she should still have more than half left. $54\%$ is just over half — exactly the right neighborhood. The answer is also independent of starting count: with $200$ marbles the same logic gives $200 \to 160 \to 144 \to 108$, and $\tfrac{108}{200} = 54\%$. Same percentage, confirming the result.
Alternative: Tool #5 (Look for a Pattern) on the "keep fractions" gives a one-line solution: at each stage Gilda keeps $\tfrac{4}{5}$, then $\tfrac{9}{10}$, then $\tfrac{3}{4}$ of what she had. Multiplying: $\tfrac{4}{5} \times \tfrac{9}{10} \times \tfrac{3}{4} = \tfrac{108}{200} = \tfrac{54}{100} = 54\%$. The pattern "each step multiplies the kept fraction" turns three subtractions into one product.
CCSS standards used (min grade 6)
5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $25\%$ of $72$ as $\tfrac{1}{4} \times 72 = 18$ — a Grade 5 unit-fraction-times-whole-number multiplication.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Tracking the percentage of marbles at each stage, where every percent is taken of the *current* bag (a shifting base) — the core Grade 6 percent-reasoning standard.)
⭐ This AMC 8 problem only needs Grade 6 percent reasoning — taking a percent of whatever is left at each step — that you already know!
⭐ This AMC 8 problem only needs Grade 6 percent reasoning — taking a percent of whatever is left at each step — that you already know!
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