2001 AMC 10 Problems/Problem 7

Problem

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?

$\textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$

Solution

If $x$ is the number, then moving the decimal point four places to the right is the same as multiplying $x$ by $10000$ . This gives us the equation $10000x=4\cdot\frac{1}{x}$ . This is equivalent to $x^2=\frac{4}{10000}$ . Since $x$ is positive, it follows that $x=\frac{2}{100}=\boxed{(C) 0.02}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 10 Problems and Solutions

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2001 AMC 10 Problems/Problem 7

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