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2005 AMC 10B Problem 3

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2005 AMC 10B Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Ratios, Percents & Averages · difficulty 1 of 5 · about 1 min · position P1-10

A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?

  1. A)

    110\frac{1}{10}

  2. B)

    19\frac{1}{9}

  3. C)

    13\frac{1}{3}

  4. D)

    49\frac{4}{9}

  5. E)

    59\frac{5}{9}

Answer

D

Key insight

Each day leaves two thirds of what was there, so after two days two thirds of two thirds, or four ninths, remains.

Solution

Using one third of a quantity leaves two thirds of it.

After day one, 23\frac{2}{3} of the gallon remains.

On day two, one third of that amount is used, so two thirds of it is kept:
23⋅23=49.\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}.

So 49\frac{4}{9} of the original gallon is available on the third day.

The answer is (D) 49\boxed{\textbf{(D)}\ \frac{4}{9}}.

Why this works

Repeated "use a fraction of what is left" is multiplicative, not additive: each step scales the remaining amount by the fraction kept. Tracking the fraction kept (23\frac{2}{3}) rather than the fraction used avoids the subtraction bookkeeping entirely.

The trap

Subtracting one third twice (1 - 1/3 - 1/3 = 1/3), forgetting that the second third is of the remaining paint.

Common mistakes

  • Subtracting one third twice (1 - 1/3 - 1/3 = 1/3), forgetting that the second third is of the remaining paint.
  • Computing the amount used on day two (29\frac{2}{9}) or the total used (59\frac{5}{9}) and stopping there.

Techniques

Set up the equation/formula and compute; no special trick needed