2011 AMC 8 Problem 2

Below is the professionally curated solution for Problem 2 of the 2011 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 8 solutions, or check the answer key.

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Concepts:arearectangle

Difficulty rating: 450

2.

Karl's rectangular vegetable garden is 2020 feet by 4545 feet, and Makenna's is 2525 feet by 4040 feet. Whose garden is larger in area?

Karl’s garden is larger by 100100 square feet.

Karl’s garden is larger by 2525 square feet.

The gardens are the same size.

Makenna’s garden is larger by 2525 square feet.

Makenna’s garden is larger by 100100 square feet.

Solution:

The area of Karl's garden is 20ft45ft=900ft2.20 \text{ft} \cdot 45 \text{ft} = 900 \text{ft}^2. The area of Makenna's garden is 25ft40ft=1000ft2.25 \text{ft} \cdot 40 \text{ft} = 1000 \text{ft}^2.

The difference of these areas is 1000ft2900ft2=100ft2.1000 \text{ft}^2 - 900 \text{ft}^2 = 100 \text{ft}^2. Therefore, Makenna's garden is 100ft2100 \text{ft}^2 larger than Karl's.

Thus, E is the correct answer.

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