Difference between revisions of "2001 AMC 10 Problems/Problem 7"
Pidigits125 (talk | contribs) (→Solution) |
|||
| (9 intermediate revisions by 6 users not shown) | |||
| Line 5: | Line 5: | ||
<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: | |
| + | <cmath>10000x=4\cdot\frac{1}{x} \implies x^2=\frac{4}{10000}</cmath> | ||
| + | Since <cmath>x>0\implies x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}</cmath> | ||
| − | + | == 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. | |
| − | + | ==Video Solution by Daily Dose of Math== | |
| − | + | https://youtu.be/M-mjlmRQByI?si=8jC0mYpCej1cRTIQ | |
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2001|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:00, 15 July 2024
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:
![\[10000x=4\cdot\frac{1}{x} \implies x^2=\frac{4}{10000}\]](http://latex.artofproblemsolving.com/9/6/7/96724db51992bc6441cf1b4147fe0f9e17a50c2e.png)
Since ![\[x>0\implies x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}\]](http://latex.artofproblemsolving.com/0/c/0/0c03437b2567c9e3f796b68890bd2646881c44e4.png)
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.
Video Solution by Daily Dose of Math
https://youtu.be/M-mjlmRQByI?si=8jC0mYpCej1cRTIQ
~Thesmartgreekmathdude
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. 
