AMC 10 · 2011 · #3
Easy mode Grade 5A store rounds every length to the nearest whole inch. So a side labeled x inches could really be anything from x−0.5 inches up to x+0.5 inches.
A rectangular tile is labeled 2 inches on one side and 3 inches on the other. What is the smallest area the tile could really have, in square inches?
Pick an answer.
AMC 10 2011 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: A store reports a length of $x$ inches to mean the true length is somewhere from $x-0.5$ inches up to $x+0.5$ inches. A rectangular tile is reported as $2$ inches by $3$ inches. Find the smallest possible area of the tile, in square inches.
Givens: A reported length $x$ means the true length is at least $x-0.5$ and at most $x+0.5$ inches; The tile's reported dimensions are $2$ inches by $3$ inches; Answer choices: (A) $3.75$, (B) $4.5$, (C) $5$, (D) $6$, (E) $8.75$
Unknowns: The minimum possible area of the rectangle, in square inches
Understand
Restated: A store reports a length of $x$ inches to mean the true length is somewhere from $x-0.5$ inches up to $x+0.5$ inches. A rectangular tile is reported as $2$ inches by $3$ inches. Find the smallest possible area of the tile, in square inches.
Givens: A reported length $x$ means the true length is at least $x-0.5$ and at most $x+0.5$ inches; The tile's reported dimensions are $2$ inches by $3$ inches; Answer choices: (A) $3.75$, (B) $4.5$, (C) $5$, (D) $6$, (E) $8.75$
Plan
Primary tool: #14 Extreme Principle
Secondary: #8 Analyze the Units, #7 Identify Subproblems
The question asks for a minimum, so Tool #14 (Extreme Principle) drives the solution: push each measurement to the smallest value its reported range allows. Tool #7 (Identify Subproblems) splits the job into two easy pieces — find the smallest each side can be, then multiply. Tool #8 (Analyze the Units) confirms that inches times inches gives square inches, so the number we compute really is an area.
Execute — Answer: A
5.NBT.B.7 Step 1 Find the smallest each side can be
- A reported length can be off by up to $0.5$ inch in either direction, so the smallest a side can truly be is $\text{(reported)} - 0.5$.
- The $2$-inch side is at least $2-0.5=1.5$ inches, and the $3$-inch side is at least $3-0.5=2.5$ inches.
💡 To make the tile as small as possible, shrink every side to the lowest value its label allows.
5.NF.B.5 Step 2 Smallest sides give the smallest area
- Area is length times width, and both are positive.
- Making either factor smaller makes the product smaller, so the area is smallest exactly when both sides are at their smallest.
- That means we pair the minimum length $1.5$ with the minimum width $2.5$ — not a mix of one small and one large side.
💡 Multiplying by a smaller number always scales the result down, so the two smallest sides give the smallest area.
4.MD.A.3 Step 3 Set up the minimum area
- Apply the rectangle area formula to the two minimum dimensions.
- The minimum area is $1.5$ inches times $2.5$ inches.
- Checking units, inches $\times$ inches gives square inches, which is what an area should be.
💡 A rectangle's area is just its two side lengths multiplied together.
5.NBT.B.7 Step 4 Multiply the decimals
- Compute $1.5\times2.5$.
- One way: $1.5\times2.5=1.5\times2+1.5\times0.5=3+0.75=3.75$.
- So the minimum area is $3.75$ square inches, which is choice (A).
- The other choices come from not shrinking the sides: $2\times3=6$ keeps both reported values (D), and $2.5\times3.5=8.75$ uses the maximum sides (E).
💡 Split the multiplication into a whole part and a half part, then add.
5.NBT.B.7 A reported length can be off by up to $0.5$ inch in either direction, so the sma 5.NF.B.5 Area is length times width, and both are positive. Making either factor smaller 4.MD.A.3 Apply the rectangle area formula to the two minimum dimensions. The minimum area 5.NBT.B.7 Compute $1.5\times2.5$. One way: $1.5\times2.5=1.5\times2+1.5\times0.5=3+0.75=3. Review
Reasonableness: The reported area would be $2\times3=6$ square inches, so the minimum must be smaller than $6$ — and $3.75<6$, which fits. It should also be larger than zero and not too far below $6$, since each side only drops by $0.5$; $3.75$ sits sensibly between the maximum area $2.5\times3.5=8.75$ and something much smaller. The trap answers $6$ and $8.75$ come from using the reported or maximum sides instead of the minimum ones.
Alternative: Test the extremes directly against the choices. The maximum area is $2.5\times3.5=8.75$ (choice E) and the reported area is $2\times3=6$ (choice D); since the question wants the minimum, both are too big, and only the smallest-sides product $1.5\times2.5=3.75$ remains, confirming (A).
CCSS standards used (min grade 5)
5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Subtracting $0.5$ from each reported side ($2-0.5$, $3-0.5$) and multiplying $1.5\times2.5=3.75$.)5.NF.B.5Interpret multiplication as scaling or resizing (Reasoning that making each factor smaller scales the product down, so minimum sides give the minimum area.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Using area $=$ length $\times$ width to turn the two minimum side lengths into the minimum area.)
⭐ To make a product as small as possible, make each thing you multiply as small as it is allowed to be.
⭐ To make a product as small as possible, make each thing you multiply as small as it is allowed to be.
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