# Difference between revisions of "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 1

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 $x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}$

## Solution 2

Alternatively, we could try each solution and see if it fits the problems given statements.

After testing, we find that $\boxed{\textbf{(C)}\ 0.02}$ is the correct answer.

~ljlbox

## See Also

 2001 AMC 10 (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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