AMC 8 · 2010 · #12

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

percentagefraction-arithmeticlinear-equations-one-var identify-subproblemsconvert-to-algebra ↑ Prerequisites: percentagefraction-arithmetic

📏 Medium solution 💡 3 insights

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Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

100

(E)

150

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find an Invariant

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.

1STEP 1

Red is 80% of 500 — that's 400 red, and the rest are 100 blue.

red = 0.80 × 500 = 400, blue = 500 - 400 = 100

2STEP 2

Only reds leave, so the blue count never changes — it stays at 100.

blue_final = blue_initial = 100

3STEP 3

If 75% of the final bag is red, the other 25% is blue — so the 100 blue balls are 25% of the new total.

0.25 × T = 100

4STEP 4

Divide both sides by 0.25 (times four): the new total is 400 balls.

T =

\frac{100}{0.25}

= 400

5STEP 5

The bag dropped from 500 to 400, and only reds left, so 100 red balls were removed.

red removed = 500 - 400 = 100 → (D)

Answer

100

Check the final bag: 400 balls total, with 100 blue and 400 - 100 = 300 red. Red fraction =

\frac{300}{400}

= 0.75 = 75%. That matches the target exactly, so removing 100 red balls is correct.

💡Key takeaway

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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