1988 AMC 8 Problem 6 — Solution | LIVE by Po-Shen Loh

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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

$\dfrac{(0.2)^3}{(0.02)^2}$

is

$0.2$

$2$

$10$

$15$

$20$

Answer: E

Concepts:decimal exponent

Difficulty rating: 730

Small Hint:

Compute the numerator and denominator separately: $(0.2)^3 = 0.008$

Big Hint:

$(0.02)^2 = 0.0004,$ then divide

Solution:

The numerator is $(0.2)^3 = 0.008$ and the denominator is $(0.02)^2 = 0.0004.$

So the value is $\dfrac{0.008}{0.0004} = \dfrac{80}{4} = 20.$

Thus, the correct answer is E .

Problem 6 in Other Years

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