Difference between revisions of "2001 AMC 12 Problems/Problem 3"
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=== Solution 2 === | === Solution 2 === | ||
| − | Let <math>A</math> be Kristin's annual income. Notice that <cmath>p\%\cdot28000 + (p + 2)\%\cdot(A - 28000)</cmath> <cmath>= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000)</cmath> <cmath>= p\%\ | + | Let <math>A</math> be Kristin's annual income. Notice that <cmath>p\%\cdot28000 + (p + 2)\%\cdot(A - 28000)</cmath> <cmath>= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000)</cmath> <cmath>= p\%\cdotA + 2\%\cdot(A - 28000)</cmath> |
== See Also == | == See Also == | ||
Revision as of 17:51, 30 June 2019
- The following problem is from both the 2001 AMC 12 #3 and 2001 AMC 10 #9, so both problems redirect to this page.
Problem
The state income tax where Kristin lives is levied at the rate of 
of the first
of annual income plus 
of any amount above . Kristin
noticed that the state income tax she paid amounted to 
of her
annual income. What was her annual income?
Solution
Solution 1
Let the income amount be denoted by 
.
We know that 
.
We can now try to solve for :
So the answer is 
Solution 2
Let be Kristin's annual income. Notice that ![\[p\%\cdot28000 + (p + 2)\%\cdot(A - 28000)\]](http://latex.artofproblemsolving.com/3/e/d/3edfcb854a8b31289355ae466d4638883cb14712.png)
![\[= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000)\]](http://latex.artofproblemsolving.com/c/c/4/cc40be936c4bde87d752370d4cffb5c674dd84a4.png)
\[= p\%\cdotA + 2\%\cdot(A - 28000)\] (Error compiling LaTeX. Unknown error_msg)
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 12 Problems and Solutions | |
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
