Difference between revisions of "2001 AMC 10 Problems/Problem 7"
Pidigits125 (talk | contribs) (→Solution) |
|||
Line 20: | Line 20: | ||
{{AMC10 box|year=2001|num-b=6|num-a=8}} | {{AMC10 box|year=2001|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Revision as of 10:10, 4 July 2013
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
We can write our equation as:
Cross-multiply and solve for .
.
.
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.