Difference between revisions of "2001 AMC 10 Problems/Problem 7"
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<math> \textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2 </math> | <math> \textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2 </math> | ||
| − | == Solution == | + | == Solution 1 == |
If <math>x</math> is the number, then moving the decimal point four places to the right is the same as multiplying <math>x</math> by <math>10000</math>. This gives us the equation | If <math>x</math> is the number, then moving the decimal point four places to the right is the same as multiplying <math>x</math> by <math>10000</math>. This gives us the equation | ||
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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{\textbf{(C)}\ 0.02}</math> | Since <math>x</math> is positive, it follows that <math>x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}</math> | ||
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| + | == Solution 2 == | ||
| + | Alternatively, we could try each solution and see if it fits the problems given statements. | ||
| + | |||
| + | After testing, we find that <math>\boxed{\textbf{(C)}\ 0.02}</math> is the correct answer. | ||
| + | |||
| + | ~ljlbox | ||
== See Also == | == See Also == | ||
Revision as of 13:24, 20 August 2019
Contents
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 1
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 
Solution 2
Alternatively, we could try each solution and see if it fits the problems given statements.
After testing, we find that 
is the correct answer.
~ljlbox
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. 
