AMC 8 · 2019 · #8
Grade 6 arithmeticProblem
Gilda has a bag of marbles. She gives of them to her friend Pedro. Then Gilda gives of what is left to another friend, Ebony. Finally, Gilda gives 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.
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!