1986 AMC 8 Problem 4

Attempt Problem 4 of the 1986 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 1986 AMC 8 solutions, or check the answer key.

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

4.

The product (1.8)(40.3+0.07)(1.8)(40.3 + 0.07) is closest to
77
4242
7474
8484
737737
Answer: C
Concepts:estimation
Small hint:
Round each factor: 1.81.8 is near 22 and 40.3+0.0740.3 + 0.07 is near 4040
Big hint:
Estimate 2×402 \times 40 and adjust slightly downward since 1.8<21.8 \lt 2
Solution:
The second factor 40.3+0.07=40.3740.3 + 0.07 = 40.37 is about 40,40, and 1.81.8 is about 2.2. A quick estimate is 2(40)−0.2(40)=80−8=72,2(40) - 0.2(40) = 80 - 8 = 72, so the product is closest to 74.74.
Thus, the correct answer is C .
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