Difference between revisions of "2001 AMC 12 Problems/Problem 3"
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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #3]] and [[2001 AMC 10 Problems|2001 AMC 10 #9]]}} | {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #3]] and [[2001 AMC 10 Problems|2001 AMC 10 #9]]}} | ||
| − | == Problem == | + | ==Problem== |
The state income tax where Kristin lives is levied at the rate of <math>p\%</math> of the first | The state income tax where Kristin lives is levied at the rate of <math>p\%</math> of the first | ||
<math>\textdollar 28000</math> of annual income plus <math>(p + 2)\%</math> of any amount above <math>\textdollar 28000</math>. Kristin | <math>\textdollar 28000</math> of annual income plus <math>(p + 2)\%</math> of any amount above <math>\textdollar 28000</math>. Kristin | ||
| Line 7: | Line 7: | ||
annual income. What was her annual income? | annual income. What was her annual income? | ||
| − | <math>\ | + | <math>\textbf{(A)}\,\textdollar 28000 \qquad \textbf{(B)}\,\textdollar 32000 \qquad \textbf{(C)}\,\textdollar 35000 \qquad \textbf{(D)}\,\textdollar 42000 \qquad \textbf{(E)}\,\textdollar 56000</math> |
| − | == Solution== | + | ==Solution 1== |
| − | |||
| − | + | Let <math>A</math>, <math>T</math> be Kristin's annual income and the income tax total, respectively. Notice that | |
| + | <cmath>\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*}</cmath> | ||
| + | We are also given that <cmath>T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A</cmath> | ||
| + | Thus, <cmath>p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A</cmath> <cmath>2\%\cdot(A - 28000) = 0.25\%\cdot A</cmath> | ||
| + | Solve for <math>A</math> to obtain <math>A = \boxed{\textbf{(B) }\textdollar 32000} </math>. | ||
| − | + | ~ Nafer | |
| − | + | ==Video Solution by Daily Dose of Math== | |
| − | + | https://youtu.be/io1JWEGl_J0?si=qa09mILGOdoNM5pc | |
| − | + | ~Thesmartgreekmathdude | |
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== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=2|num-a=4}} | {{AMC12 box|year=2001|num-b=2|num-a=4}} | ||
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{{AMC10 box|year=2001|num-b=8|num-a=10}} | {{AMC10 box|year=2001|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:40, 15 July 2024
- 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 1
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
Video Solution by Daily Dose of Math
https://youtu.be/io1JWEGl_J0?si=qa09mILGOdoNM5pc
~Thesmartgreekmathdude
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. 
