2011 AMC 8 Problem 2

Attempt Problem 2 of the 2011 AMC 8 below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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.
Answer: E
Concepts:arearectangle
Small hint:
Compare 20⋅4520\cdot45 with 25⋅4025\cdot40.
Big hint:
Compute each rectangular area separately.
Solution:
The area of Karl’s garden is 20ft⋅45ft=900ft2.20 \text{ft} \cdot 45 \text{ft} = 900 \text{ft}^2. The area of Makenna’s garden is 25ft⋅40ft=1000ft2.25 \text{ft} \cdot 40 \text{ft} = 1000 \text{ft}^2.
The difference of these areas is 1000ft2−900ft2=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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