2009 AMC 8 Problem 8

Below is the professionally curated solution for Problem 8 of the 2009 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 8 solutions, or check the answer key.

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Concepts:areapercentage

Difficulty rating: 920

8.

The length of a rectangle is increased by 10%10\% and the width is decreased by 10%10\%. What percent of the old area is the new area?

9090

9999

100100

101101

110110

Solution:

Let the old length and width be ll and ww, so the old area is lwlw.

The new length is 1.1l1.1l, and the new width is 0.9w0.9w. Thus the new area is 1.10.9lw=0.99lw1.1\cdot0.9lw=0.99lw.

This shows that the new area is 99%99\% of the old area.

Thus, B is the correct answer.

Problem 8 in Other Years

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