AMC 10 · 2009 · #3

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

ratio-proportionrate dimensional-analysis ↑ Prerequisites: ratio-proportion

📏 Short solution 💡 2 insights

Problem

Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?

Pick an answer.

(A)

$\ 10$

(B)

$\ 12$

(C)

$\ 15$

(D)

$\ 18$

(E)

$\ 25$

AMC 10 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: Paula's paint supply is enough for exactly 30 identical rooms. She loses 3 cans, and the paint that is left is enough for exactly 25 rooms. Find how many cans she uses to paint those 25 rooms.

Givens: The full supply paints exactly $30$ rooms; After $3$ cans fall off the truck, the remaining paint is enough for exactly $25$ rooms; Every room is identical in size and uses the same amount of paint; Answer choices: (A) $10$, (B) $12$, (C) $15$, (D) $18$, (E) $25$

Unknowns: The number of cans of paint used for the 25 rooms

Understand

Plan

Primary tool: #8 Analyze the Units

Secondary: #7 Identify Subproblems

The whole problem hinges on one hidden rate: how much paint one room takes, measured in cans per room. Tool #8 (Analyze the Units) puts that rate at the center — once we know cans per room, the 25-room total is one multiplication. Tool #7 (Identify Subproblems) sets it up in stages: first turn the lost cans into a number of lost rooms, then find the per-room rate, then scale up to 25 rooms.

Execute — Answer: C

#7 Identify Subproblems 4.OA.A.3 Step 1

Turn the lost cans into lost rooms

The full supply covers $30$ rooms and what is left covers $25$ rooms, so the paint that fell off covers $30-25=5$ rooms.
That missing paint is exactly the $3$ cans.
So $3$ cans is the same amount of paint as $5$ rooms need.

$$30-25=5\text{ rooms},\qquad 3\text{ cans} = 5\text{ rooms of paint}$$

💡 Whatever was lost is just the gap between what she could paint before and after.

#8 Analyze the Units 6.RP.A.2 Step 2

Find the paint per room

If $3$ cans cover $5$ rooms, then one room takes $3$ cans shared over $5$ rooms, which is $\dfrac{3}{5}$ of a can.
This is the unit rate: cans per room.

$$\dfrac{3\text{ cans}}{5\text{ rooms}}=\dfrac{3}{5}\text{ can per room}$$

💡 Splitting the cans evenly among the rooms they cover gives the cost of a single room.

#8 Analyze the Units 5.NF.B.4 Step 3

Scale up to 25 rooms

To paint $25$ rooms she needs $25$ times the per-room amount: $25\times\dfrac{3}{5}$.
Since $\dfrac{25}{5}=5$, this is $5\times3=15$ cans.
So Paula used $15$ cans for the 25 rooms, which is choice (C).

$$25\times\dfrac{3}{5}=\dfrac{25}{5}\times3=5\times3=15\;\Rightarrow\;\textbf{(C)}$$

💡 Once you know the cost of one room, any number of rooms is just repeated copies of that cost.

[1] #7 4.OA.A.3 The full supply covers $30$ rooms and what is left covers $25$ rooms, so the pai

[2] #8 6.RP.A.2 If $3$ cans cover $5$ rooms, then one room takes $3$ cans shared over $5$ rooms,

[3] #8 5.NF.B.4 To paint $25$ rooms she needs $25$ times the per-room amount: $25\times\dfrac{3}

Review

Reasonableness: Check the whole story: at $\tfrac{3}{5}$ can per room, the full 30-room supply is $30\times\tfrac{3}{5}=18$ cans. Drop the $3$ lost cans and $18-3=15$ cans remain — exactly enough for $25\times\tfrac{3}{5}=15$ cans of work. Everything lines up, so $15$ is consistent. The trap answer $18$ (D) is the full original supply, not what she used on the 25 rooms; $25$ (E) just echoes the number of rooms.

Alternative: Use a variable for the per-room amount. Let $r$ be cans per room. The full supply is $30r$ cans and the reduced supply is $25r$ cans, and they differ by the $3$ lost cans: $30r-25r=3$, so $5r=3$ and $r=\tfrac{3}{5}$. Then the 25 rooms use $25r=25\times\tfrac{3}{5}=15$ cans, again (C).

CCSS standards used (min grade 6)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Subtracting $30-25=5$ to see that the 3 lost cans equal the paint for 5 rooms.)
6.RP.A.2 Understand the concept of a unit rate and use rate language (Turning "3 cans for 5 rooms" into the unit rate of $\dfrac{3}{5}$ can per room.)
5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Computing $25\times\dfrac{3}{5}=15$ to get the cans used for 25 rooms.)

⭐ Find how much paint one room needs first; then any number of rooms is just that amount multiplied up.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Turn the lost cans into lost rooms

Find the paint per room

Scale up to 25 rooms

Review

CCSS standards used (min grade 6)

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