1986 AMC 8 Problem 4

Below is the professionally curated solution for Problem 4 of the 1986 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1986 AMC 8 solutions, or check the answer key.

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

Concepts:estimation

Difficulty rating: 730

The product $(1.8)(40.3 + .07)$ is closest to

$7$

$42$

$74$

$84$

$737$

Solution:

The second factor $40.3 + .07 = 40.37$ is about $40,$ and $1.8$ is about $2.$ A quick estimate is $2(40) - .2(40) = 80 - 8 = 72,$ so the product is closest to $74.$

Thus, the correct answer is C .

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