2001 AMC 10 Problem 7

Below is the professionally curated solution for Problem 7 of the 2001 AMC 10, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 10 solutions, or check the answer key.

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Concepts:place value decimal algebraic manipulation

Difficulty rating: 1170

When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$0.0002$

$0.002$

$0.02$

$0.2$

$2$

Solution:

Moving the decimal four places right multiplies $x$ by $10000.$ So $10000x=4\cdot\dfrac1x,$ giving $x^2=\dfrac{4}{10000}.$

Since $x\gt0,$ $x=\dfrac{2}{100}=0.02.$

Thus, the correct answer is C.

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