Difference between revisions of "2001 AMC 12 Problems/Problem 3"
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<math>A=32000</math> | <math>A=32000</math> | ||
| − | So the answer is <math>\boxed{\textbf{(B) } | + | So the answer is <math>\boxed{\textbf{(B) }32000} </math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 09:37, 8 November 2021
- The following problem is from both the 2001 AMC 12 #3 and 2001 AMC 10 #9, so both problems redirect to this page.
Contents
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 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 and the income tax total, respectively. Notice that
![\begin{align*} T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\ &= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\ &= p\%\cdot A + 2\%\cdot(A - 28000) \end{align*}](http://latex.artofproblemsolving.com/a/3/9/a394ead7d83dc9889466a806e206ea2bbdcc9545.png)
We are also given that ![\[T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A\]](http://latex.artofproblemsolving.com/7/5/4/75405b5351821fc58d14290fd4dddf42a77ec4da.png)
Thus, ![\[p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A\]](http://latex.artofproblemsolving.com/6/f/b/6fb9c5d80e3f1733bd16e4837f9bcbeb5a8e7150.png)
![\[2\%\cdot(A - 28000) = 0.25\%\cdot A\]](http://latex.artofproblemsolving.com/d/1/0/d1025839a18cab02a12511105eb85edae3851ec6.png)
Solve for to obtain 
. 
~ Nafer
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. 
