AMC 10 · 2004 · #14
Grade 6 rate-ratioA bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only 31 of the marbles in the bag are blue. Then yellow marbles are added to the bag until only 51 of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?
Pick an answer.
AMC 10 2004 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: A bag holds red and blue marbles, with more blue than red. First, red marbles are added until blue marbles are $\frac{1}{3}$ of the bag. Next, yellow marbles are added until blue marbles are $\frac{1}{5}$ of the bag. Finally, the number of blue marbles is doubled. Find what fraction of the marbles are blue at the end.
Givens: The bag starts with only red and blue marbles, and there are more blue than red.; Red marbles are added until blue marbles make up $\frac{1}{3}$ of the bag.; Then yellow marbles are added until blue marbles make up $\frac{1}{5}$ of the bag.; Finally the number of blue marbles is doubled.; Answer choices: (A) $\frac{1}{5}$, (B) $\frac{1}{4}$, (C) $\frac{1}{3}$, (D) $\frac{2}{5}$, (E) $\frac{1}{2}$.
Unknowns: The fraction of the marbles that are blue after the blue marbles are doubled.
Understand
Restated: A bag holds red and blue marbles, with more blue than red. First, red marbles are added until blue marbles are $\frac{1}{3}$ of the bag. Next, yellow marbles are added until blue marbles are $\frac{1}{5}$ of the bag. Finally, the number of blue marbles is doubled. Find what fraction of the marbles are blue at the end.
Givens: The bag starts with only red and blue marbles, and there are more blue than red.; Red marbles are added until blue marbles make up $\frac{1}{3}$ of the bag.; Then yellow marbles are added until blue marbles make up $\frac{1}{5}$ of the bag.; Finally the number of blue marbles is doubled.; Answer choices: (A) $\frac{1}{5}$, (B) $\frac{1}{4}$, (C) $\frac{1}{3}$, (D) $\frac{2}{5}$, (E) $\frac{1}{2}$.
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #4 Introduce a Variable, #9 Solve an Easier Related Problem
The last action is doubling the blue marbles, so the only snapshot that decides the answer is the bag right before that: blue marbles are $\frac{1}{5}$ of it. Everything earlier — the $\frac{1}{3}$ stage, the starting counts — is a distraction that never touches the final fraction, which is why Tool #9 (Solve an Easier Related Problem) lets us throw it away. The winning move is Tool #16 (Change Focus / Count the Complement): stop chasing the blue count and watch the non-blue marbles, because doubling blue leaves them untouched. Tool #4 (Introduce a Variable) turns the $\frac{1}{5}$ into concrete counts so the doubling is easy to track.
Execute — Answer: C
6.RP.A.3 Step 1 Keep only the last snapshot
- The final action is doubling the blue marbles, so the answer depends only on the bag as it stands just before that doubling.
- At that moment blue marbles are $\frac{1}{5}$ of the bag.
- The earlier step where blue was $\frac{1}{3}$ only helped set up counts along the way; it does not change the fraction after the doubling, so it can be ignored.
💡 Only the state right before the last move can affect the result, so the rest is noise.
6.RP.A.1 Step 2 Split the bag into 5 equal parts
- Blue is $\frac{1}{5}$ of the bag, so picture the bag as $5$ equal parts: $1$ part blue and $4$ parts non-blue.
- Let each part hold $b$ marbles.
- Then there are $b$ blue marbles and $4b$ non-blue marbles, for a total of $5b$.
💡 Turning $\frac{1}{5}$ into $1$ part out of $5$ makes the blue-to-rest split concrete.
3.OA.C.7 Step 3 Double the blue, leave the rest alone
- Doubling the blue marbles turns $b$ into $2b$.
- Adding blue marbles does nothing to the non-blue marbles, so they stay at $4b$.
- The new total is $2b+4b=6b$.
💡 The non-blue pile is a fixed anchor — doubling blue changes blue and the total, not it.
6.RP.A.3 Step 4 Recount the blue fraction
- Now the bag has $2b$ blue marbles out of $6b$ total, so the fraction that is blue is $\frac{2b}{6b}=\frac{2}{6}=\frac{1}{3}$.
- The answer is (C).
💡 With the non-blue part fixed, $1$ part blue out of $5$ becomes $2$ parts out of $6$.
6.RP.A.3 The final action is doubling the blue marbles, so the answer depends only on the 6.RP.A.1 Blue is $\frac{1}{5}$ of the bag, so picture the bag as $5$ equal parts: $1$ par 3.OA.C.7 Doubling the blue marbles turns $b$ into $2b$. Adding blue marbles does nothing 6.RP.A.3 Now the bag has $2b$ blue marbles out of $6b$ total, so the fraction that is blu Review
Reasonableness: Put in real numbers with $b=10$: start with $10$ blue and $40$ non-blue, a total of $50$, and indeed $\frac{10}{50}=\frac{1}{5}$ is blue. Doubling gives $20$ blue and still $40$ non-blue, total $60$, and $\frac{20}{60}=\frac{1}{3}$ — choice (C). Notice the trap: doubling a $\frac{1}{5}$ share does not give $\frac{2}{5}$ (choice (D)), because the total grows too. The blue share rises from $\frac{1}{5}$ to $\frac{1}{3}$, a modest increase, which fits the fact that blue was still a minority after doubling.
Alternative: Run the whole story with numbers to confirm the ignored steps really don't matter. Start with $2$ blue and $1$ red (more blue than red). Add red until blue is $\frac{1}{3}$: the total must be $6$, so add $3$ red, giving $2$ blue and $4$ red. Add yellow until blue is $\frac{1}{5}$: the total must be $10$, so add $4$ yellow, giving $2$ blue out of $10$. Double the blue: $4$ blue out of $12$, and $\frac{4}{12}=\frac{1}{3}$ — the same answer (C).
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Recognizing that only the $\frac{1}{5}$-blue snapshot decides the answer, and computing the final blue fraction $\frac{2}{6}=\frac{1}{3}$.)6.RP.A.1Understand the concept of a ratio and use ratio language (Reading $\frac{1}{5}$ blue as the part-to-whole split $1$ part blue to $4$ parts non-blue and assigning $b$, $4b$, $5b$.)3.OA.C.7Fluently multiply and divide within 100 (Doubling the blue count from $b$ to $2b$ and adding to find the new total $6b$.)
⭐ Doubling the blue marbles does not double their fraction — the non-blue pile stays put, so $1$ part blue out of $5$ becomes $2$ parts out of $6$, which is $\frac{1}{3}$.
⭐ Doubling the blue marbles does not double their fraction — the non-blue pile stays put, so $1$ part blue out of $5$ becomes $2$ parts out of $6$, which is $\frac{1}{3}$.
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