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Difference between revisions of "2001 AMC 10 Problems/Problem 7"

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== Problem ==
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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?
 
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?
  
 
<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>
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== Solution 1 ==
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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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<cmath>10000x=4\cdot\frac{1}{x}</cmath>
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This is equivalent to <cmath>x^2=\frac{4}{10000}</cmath>
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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 ==
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Alternatively, we could try each solution and see if it fits the problems given statements.
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After testing, we find that <math>\boxed{\textbf{(C)}\ 0.02}</math> is the correct answer.
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~ljlbox
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== See Also ==
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{{AMC10 box|year=2001|num-b=6|num-a=8}}
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{{MAA Notice}}

Revision as of 14:24, 20 August 2019

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 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

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