AMC 8 · 2009 · #8
Grade 6 geometry-2dProblem
The length of a rectangle is increased by 10% percent and the width is decreased by 10% percent. What percent of the old area is the new area?
Pick an answer.
AMC 8 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.
Toolkit + CCSS Solution
Understand
Restated: A rectangle's length grows by $10\%$ and its width shrinks by $10\%$. The new area is what percent of the old area?
Givens: Original rectangle has length $L$ and width $W$, so old area $= L \times W$; New length is $10\%$ larger than $L$; New width is $10\%$ smaller than $W$; Answer choices: (A) $90$, (B) $99$, (C) $100$, (D) $101$, (E) $110$
Unknowns: The new area expressed as a percent of the old area
Understand
Restated: A rectangle's length grows by $10\%$ and its width shrinks by $10\%$. The new area is what percent of the old area?
Givens: Original rectangle has length $L$ and width $W$, so old area $= L \times W$; New length is $10\%$ larger than $L$; New width is $10\%$ smaller than $W$; Answer choices: (A) $90$, (B) $99$, (C) $100$, (D) $101$, (E) $110$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #1 Draw a Diagram
The answer doesn't depend on the actual size of the rectangle, so Tool #9 (Solve an Easier Related Problem) says: replace $L$ and $W$ with friendly numbers — pick $L = W = 10$ — and compute both areas directly. Tool #1 (Draw a Diagram) keeps the picture honest: sketch the $10 \times 10$ rectangle, then the new $11 \times 9$ rectangle, and compare areas by counting. This skips the algebra route (Tool #13) entirely and shows why $+10\%$ and $-10\%$ don't cancel.
Execute — Answer: B
4.MD.A.3 Step 1 - Pick easy numbers.
- Let $L = 10$ and $W = 10$, so the original rectangle is a $10 \times 10$ square.
- The old area is easy to read off.
💡 Using $100$ as the old area is a deliberate trick: any final answer in "percent" becomes the new area itself, no extra division needed.
6.RP.A.3 Step 2 - Apply each percent change.
- $10\%$ of $10$ is $1$, so the new length is $10 + 1 = 11$ and the new width is $10 - 1 = 9$.
💡 Draw the new $11 \times 9$ rectangle next to the original $10 \times 10$ square — one side got a little longer, the other a little shorter.
4.MD.A.3 Step 3 Compute the new area directly from the new sides.
💡 An $11 \times 9$ grid has $99$ unit squares — one less than the original $100$, even though both sides only changed by $1$.
6.RP.A.3 Step 4 - Express the new area as a percent of the old area.
- Since the old area is exactly $100$, the new area $99$ is already in "percent" form.
💡 The whole reason we picked $L = W = 10$ was to make this last conversion free.
4.MD.A.3 Pick easy numbers. Let $L = 10$ and $W = 10$, so the original rectangle is a $10 6.RP.A.3 Apply each percent change. $10\%$ of $10$ is $1$, so the new length is $10 + 1 = 4.MD.A.3 Compute the new area directly from the new sides. 6.RP.A.3 Express the new area as a percent of the old area. Since the old area is exactly Review
Reasonableness: The change is small: one side grew by $10\%$, the other shrank by $10\%$, so the area must stay near $100\%$. Choices $90\%$ and $110\%$ are far too extreme, $100\%$ would mean the changes cancel exactly (they don't, because the bigger side now multiplies by a smaller side), and $101\%$ would mean area grew. The only plausible answer is just below $100\%$ — namely $99\%$, matching (B).
Alternative: Tool #13 (Use Algebra) confirms it for any rectangle: $A_{\text{new}} = (1.1 L)(0.9 W) = (1.1 \times 0.9) L W = 0.99 \, A_{\text{old}}$. The product $1.1 \times 0.9 = 0.99$ is the same no matter what $L$ and $W$ are, so $A_{\text{new}}$ is always $99\%$ of $A_{\text{old}}$ — the easier-problem answer generalizes.
CCSS standards used (min grade 6)
4.MD.A.3Apply the area and perimeter formulas for rectangles (Computing the old area $10 \times 10 = 100$ and the new area $11 \times 9 = 99$ from length times width.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Verifying the general result with the decimal product $1.1 \times 0.9 = 0.99$ in the alternative algebraic check.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Translating the $10\%$ increase and $10\%$ decrease into the new dimensions $11$ and $9$, and expressing the area ratio $99/100$ as a percent.)
⭐ This AMC 8 problem only needs Grade 6 percent reasoning — and a clever Grade 4 trick: pick $L = W = 10$ so the old area is exactly $100$ and the answer falls out!
⭐ This AMC 8 problem only needs Grade 6 percent reasoning — and a clever Grade 4 trick: pick $L = W = 10$ so the old area is exactly $100$ and the answer falls out!
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