Difference between revisions of "2001 AMC 10 Problems/Problem 7"
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<cmath>10000x=4\cdot\frac{1}{x}</cmath> | <cmath>10000x=4\cdot\frac{1}{x}</cmath> | ||
This is equivalent to <cmath>x^2=\frac{4}{10000}</cmath> | This is equivalent to <cmath>x^2=\frac{4}{10000}</cmath> | ||
| − | Since <math>x</math> is positive, it follows that <math>x=\frac{2}{100}=\boxed{(C) 0.02}</math> | + | Since <math>x</math> is positive, it follows that <math>x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}</math> |
== See Also == | == See Also == | ||
Revision as of 22:05, 27 July 2016
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?
Solution
If 
is the number, then moving the decimal point four places to the right is the same as multiplying by 
. This gives us the equation
![\[10000x=4\cdot\frac{1}{x}\]](http://latex.artofproblemsolving.com/5/b/4/5b4337286e66f4a312e0b91b86d77766543858d2.png)
This is equivalent to ![\[x^2=\frac{4}{10000}\]](http://latex.artofproblemsolving.com/f/4/1/f415448b415435ec890acadacbbbcfeb039ee01d.png)
Since is positive, it follows that 
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| 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. 
