AMC 8 · 2009 · #8

Grade 6 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

area-rectanglespercentageratio-proportion easier-related-problem ↑ Prerequisites: area-rectanglespercentage

📏 Short solution 💡 2 insights

Problem

A rectangle's length grows by 10% and its width shrinks by 10%. The new area is what percent of the old area?

Pick an answer.

(A)

(B)

(C)

100

(D)

101

(E)

110

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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 × 10 rectangle, then the new 11 × 9 rectangle, and compare areas by counting. This skips the algebra route (Tool #13) entirely and shows why +10% and -10% don't cancel.

1STEP 1

Pick friendly numbers: let L = 10 and W = 10, so the old area is 100.

A_old = 10 × 10 = 100

2STEP 2

10% of 10 is 1, so the new length is 11 and the new width is 9.

L_new = 11, W_new = 9

3STEP 3

The new area is just 11 × 9 = 99 — one less than the old 100.

A_new = 11 × 9 = 99

4STEP 4

Since the old area is exactly 100, the new area 99 is already the percent: 99%.

A_new/A_old =

\frac{99}{100}

= 99% → (B)

Answer

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).

💡Key takeaway

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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