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 ==
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== Solution 1 ==
  
We can write our equation as:
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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:
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<cmath>10000x=4\cdot\frac{1}{x} \implies x^2=\frac{4}{10000}</cmath>
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Since <cmath>x>0\implies x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}</cmath>
  
<math> \frac{x}{10000}=\frac{4}{x} </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.
  
Cross-multiply and solve for <math> x </math>.
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After testing, we find that <math>\boxed{\textbf{(C)}\ 0.02}</math> is the correct answer.
 
 
<math> x^2=40000 \implies x=200 </math>.
 
 
 
<math> 200/10000=2/100= \boxed{\textbf{(C) }0.02} </math>.
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:25, 28 July 2022

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: \[10000x=4\cdot\frac{1}{x} \implies x^2=\frac{4}{10000}\] Since \[x>0\implies x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}\]

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.

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

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