2001 AMC 10 Problems/Problem 7

Revision as of 22:05, 27 July 2016 by 2002kevinhuang (talk | contribs) (Solution)

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

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 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. AMC logo.png

Invalid username
Login to AoPS