AMC 8 · 2010 · #12
Easy mode Grade 6Problem
Picture a large bag holding balls. 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 of the balls still in the bag are red.
How many red balls do you take out?
Pick an answer.
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.