1988 AMC 8 Problem 6

Attempt Problem 6 of the 1988 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 1988 AMC 8 solutions, or check the answer key.

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

6.

The value of
(0.2)3(0.02)2\dfrac{(0.2)^3}{(0.02)^2}
is
0.20.2
22
1010
1515
2020
Answer: E
Concepts:decimalexponent
Small hint:
Compute the numerator and denominator separately: (0.2)3=0.008(0.2)^3 = 0.008
Big hint:
(0.02)2=0.0004,(0.02)^2 = 0.0004, then divide
Solution:
The numerator is (0.2)3=0.008(0.2)^3 = 0.008 and the denominator is (0.02)2=0.0004.(0.02)^2 = 0.0004.
So the value is 0.0080.0004=804=20.\dfrac{0.008}{0.0004} = \dfrac{80}{4} = 20.
Thus, the correct answer is E .
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