AMC 8 · 2010 · #12
Easy mode Grade 6Problem
Picture a large bag holding 500 balls. 80% of them are red, and the rest are blue.
You start taking red balls out of the bag, one at a time. The blue balls stay where they are.
You want to stop the moment 75% of the balls still in the bag are red.
How many red balls do you take out?
Pick an answer.
AMC 8 2010 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 has $500$ balls. $80\%$ are red and the other $20\%$ are blue. We only remove red balls (never blue). How many red balls must be removed so that, in the bag that remains, $75\%$ of the balls are red?
Givens: Total balls $= 500$; Red fraction at start $= 80\%$, so $0.80 \times 500 = 400$ red balls; Blue fraction at start $= 20\%$, so $100$ blue balls; Only red balls are removed; the count of blue balls stays at $100$; Target: red fraction $= 75\%$ of the new total; Answer choices: (A) $25$, (B) $50$, (C) $75$, (D) $100$, (E) $150$
Unknowns: The number of red balls that must be removed
Understand
Restated: A bag has $500$ balls. $80\%$ are red and the other $20\%$ are blue. We only remove red balls (never blue). How many red balls must be removed so that, in the bag that remains, $75\%$ of the balls are red?
Givens: Total balls $= 500$; Red fraction at start $= 80\%$, so $0.80 \times 500 = 400$ red balls; Blue fraction at start $= 20\%$, so $100$ blue balls; Only red balls are removed; the count of blue balls stays at $100$; Target: red fraction $= 75\%$ of the new total; Answer choices: (A) $25$, (B) $50$, (C) $75$, (D) $100$, (E) $150$
Plan
Primary tool: #5 Find an Invariant
Secondary: #1 Draw a Picture
The key move is noticing what does NOT change: only red balls are removed, so the $100$ blue balls are the invariant. Tool #5 (Find an Invariant) reframes the problem — instead of chasing the shrinking red count, anchor on blue. If red ends at $75\%$, blue must end at $25\%$, and that $25\%$ equals the unchanged $100$ blue balls. From there, the new total is forced. Tool #1 (Draw a Picture) — a simple bar split into $75\%$ red and $25\%$ blue — makes the proportion visible at a glance.
Execute — Answer: D
6.RP.A.3 Step 1 - Count the red and blue balls at the start.
- Red is $80\%$ of $500$ and blue is the rest.
💡 Finding a percent of a whole is a Grade 6 ratio and percent skill.
6.RP.A.3 Step 2 - Identify the invariant.
- The problem only removes red balls, so the blue count never changes.
- Blue stays at $100$ no matter how many reds are removed.
💡 Locking onto the quantity that does not change is the heart of Tool #5.
6.RP.A.3 Step 3 - Translate the target.
- If $75\%$ of the final bag is red, then $100\% - 75\% = 25\%$ is blue.
- So those $100$ blue balls must be $25\%$ of the new total $T$.
💡 A bar split $75\%$/$25\%$ makes it clear blue is exactly one-quarter of the new total.
6.EE.B.7 Step 4 Solve for the new total $T$ by dividing both sides by $0.25$ (equivalently, multiply by $4$).
💡 Solving a one-step equation $0.25T = 100$ is a Grade 6 equations standard.
6.EE.B.7 Step 5 - Find how many balls were removed.
- The bag went from $500$ balls down to $400$ balls, and only red balls left, so $500 - 400 = 100$ red balls were removed.
💡 Subtracting the new total from the old total gives the number of balls taken out.
6.RP.A.3 Count the red and blue balls at the start. Red is $80\%$ of $500$ and blue is th 6.RP.A.3 Identify the invariant. The problem only removes red balls, so the blue count ne 6.RP.A.3 Translate the target. If $75\%$ of the final bag is red, then $100\% - 75\% = 25 6.EE.B.7 Solve for the new total $T$ by dividing both sides by $0.25$ (equivalently, mult 6.EE.B.7 Find how many balls were removed. The bag went from $500$ balls down to $400$ ba Review
Reasonableness: Check the final bag: $400$ balls total, with $100$ blue and $400 - 100 = 300$ red. Red fraction $= 300 / 400 = 0.75 = 75\%$. That matches the target exactly, so removing $100$ red balls is correct.
Alternative: Tool #2 (Set Up an Equation) directly: let $x$ be the red balls removed. Then red becomes $400 - x$, blue stays at $100$, and total becomes $500 - x$. The equation $\dfrac{400 - x}{500 - x} = \dfrac{3}{4}$ gives $4(400 - x) = 3(500 - x)$, so $1600 - 4x = 1500 - 3x$, hence $x = 100$ — same answer (D).
CCSS standards used (min grade 6)
6.RP.A.3Find a percent of a quantity as a rate per 100; solve problems involving finding the whole given a part and the percent (Computing $80\%$ of $500$ for the initial red count, and using "$100$ blue balls $= 25\%$ of the new total" to find the new total.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Solving the one-step equation $0.25 \times T = 100$ for the new total $T = 400$, then subtracting to find the number removed.)
⭐ When only one color is being taken out, lock onto the color that stays — the blue balls here — and the rest of the problem solves itself.
⭐ When only one color is being taken out, lock onto the color that stays — the blue balls here — and the rest of the problem solves itself.
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