AMC 10 · 2005 · #3
Easy mode Grade 5You start with one full gallon of paint. On the first day, you use one third of it. On the second day, you use one third of the paint that is still left (not one third of the whole gallon). What fraction of the original gallon is still available on the third day?
Pick an answer.
AMC 10 2005 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: One gallon of paint is used up over three days. On day 1, one third of the paint is used. On day 2, one third of whatever paint is left is used. Find what fraction of the original gallon is still available on day 3.
Givens: The starting amount is $1$ whole gallon; Day 1 uses $\dfrac{1}{3}$ of the paint; Day 2 uses $\dfrac{1}{3}$ of the paint that remains after day 1 (not $\dfrac{1}{3}$ of the original); Answer choices: (A) $\dfrac{1}{10}$, (B) $\dfrac{1}{9}$, (C) $\dfrac{1}{3}$, (D) $\dfrac{4}{9}$, (E) $\dfrac{5}{9}$
Unknowns: The fraction of the original gallon still available at the start of day 3
Understand
Restated: One gallon of paint is used up over three days. On day 1, one third of the paint is used. On day 2, one third of whatever paint is left is used. Find what fraction of the original gallon is still available on day 3.
Givens: The starting amount is $1$ whole gallon; Day 1 uses $\dfrac{1}{3}$ of the paint; Day 2 uses $\dfrac{1}{3}$ of the paint that remains after day 1 (not $\dfrac{1}{3}$ of the original); Answer choices: (A) $\dfrac{1}{10}$, (B) $\dfrac{1}{9}$, (C) $\dfrac{1}{3}$, (D) $\dfrac{4}{9}$, (E) $\dfrac{5}{9}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
The paint changes one day at a time, so Tool #7 (Identify Subproblems) handles it as two small steps: find what is left after day 1, then find what day 2 removes. Tool #16 (Change Focus / Count the Complement) keeps the eye on what is LEFT rather than what is used — the whole trap of the problem, since the paint USED totals $\dfrac{5}{9}$, which is choice (E). Tool #3 (Eliminate Possibilities) then rules out the copy-cat traps: $\dfrac{1}{9}$ if you wrongly take $\dfrac{1}{3}$ of $\dfrac{1}{3}$, and $\dfrac{1}{3}$ if you stop after day 1.
Execute — Answer: D
5.NF.A.1 Step 1 Paint left after day 1
- Start with $1$ whole gallon.
- Day 1 uses $\dfrac{1}{3}$ of it, so the paint remaining is $1-\dfrac{1}{3}$.
- Write $1$ as $\dfrac{3}{3}$ to subtract: $\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}$.
- So $\dfrac{2}{3}$ of a gallon is left when day 2 begins.
💡 Using one of three equal parts leaves the other two, so two thirds remain.
5.NF.B.4 Step 2 Paint used on day 2
- Day 2 uses $\dfrac{1}{3}$ of what is LEFT, which is $\dfrac{1}{3}$ of $\dfrac{2}{3}$ — not $\dfrac{1}{3}$ of the whole gallon.
- The word "of" here means multiply: $\dfrac{1}{3}\times\dfrac{2}{3}=\dfrac{1\times2}{3\times3}=\dfrac{2}{9}$.
- So day 2 removes $\dfrac{2}{9}$ of the original gallon.
💡 Taking a fraction "of" another fraction means multiplying the two together.
5.NF.A.1 Step 3 Paint available on day 3
- The paint available on day 3 is what was there at the start of day 2 minus what day 2 used: $\dfrac{2}{3}-\dfrac{2}{9}$.
- Rewrite $\dfrac{2}{3}$ as $\dfrac{6}{9}$ so the ninths match: $\dfrac{6}{9}-\dfrac{2}{9}=\dfrac{4}{9}$.
- That is choice (D).
- Notice the paint USED is $\dfrac{1}{3}+\dfrac{2}{9}=\dfrac{3}{9}+\dfrac{2}{9}=\dfrac{5}{9}$, which is the trap answer (E); the question asks for what is left, its complement, so $\dfrac{4}{9}$ is correct.
💡 What is left plus what is used must fill the whole gallon, so track one and the other follows.
5.NF.A.1 Start with $1$ whole gallon. Day 1 uses $\dfrac{1}{3}$ of it, so the paint remai 5.NF.B.4 Day 2 uses $\dfrac{1}{3}$ of what is LEFT, which is $\dfrac{1}{3}$ of $\dfrac{2} 5.NF.A.1 The paint available on day 3 is what was there at the start of day 2 minus what Review
Reasonableness: Check with the complement: used $=\dfrac{5}{9}$ and left $=\dfrac{4}{9}$, and $\dfrac{5}{9}+\dfrac{4}{9}=\dfrac{9}{9}=1$ whole gallon — the two parts fill exactly one gallon, so the split is consistent. The leftover $\dfrac{4}{9}\approx0.44$ is a bit less than half, which fits: after taking a third, then a third of the rest, more than half the gallon is gone but not much more. Choice (E) $\dfrac{5}{9}$ is the amount used, (C) $\dfrac{1}{3}$ stops after day 1, and (B) $\dfrac{1}{9}$ comes from multiplying $\dfrac{1}{3}\times\dfrac{1}{3}$ instead of $\dfrac{1}{3}\times\dfrac{2}{3}$.
Alternative: Track the leftover as a running product. Each day that removes $\dfrac{1}{3}$ leaves $\dfrac{2}{3}$ of what was there. After day 1: $\dfrac{2}{3}$. After day 2: $\dfrac{2}{3}\times\dfrac{2}{3}=\dfrac{4}{9}$. This lands on (D) in one multiplication, with no subtraction at all.
CCSS standards used (min grade 5)
5.NF.A.1Add and subtract fractions with unlike denominators (Finding the leftover $1-\dfrac{1}{3}=\dfrac{2}{3}$ and $\dfrac{2}{3}-\dfrac{2}{9}=\dfrac{4}{9}$ by rewriting to a common denominator.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Computing $\dfrac{1}{3}$ of the remaining $\dfrac{2}{3}$ as $\dfrac{1}{3}\times\dfrac{2}{3}=\dfrac{2}{9}$.)
⭐ "A third of what's left" means multiply by the leftover, not the whole — and since the question asks what's still there, track what stays, not what's gone.
⭐ "A third of what's left" means multiply by the leftover, not the whole — and since the question asks what's still there, track what stays, not what's gone.
More like this
Same archetype — closest grade level first.