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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]]}}
  
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== Problem ==
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==Problem==
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The state income tax where Kristin lives is levied at the rate of <math>p%</math> of the first
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The state income tax where Kristin lives is levied at the rate of <math>p\%</math> of the first
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<dollar/><math>28000</math> of annual income plus <math>(p + 2)%</math> of any amount above <dollar/><math>28000</math>. Kristin
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<math>\textdollar 28000</math> of annual income plus <math>(p + 2)\%</math> of any amount above <math>\textdollar 28000</math>. Kristin
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noticed that the state income tax she paid amounted to <math>(p + 0.25)%</math> of her
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noticed that the state income tax she paid amounted to <math>(p + 0.25)\%</math> of her
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annual income. What was her annual income?
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annual income. What was her annual income?  
  
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<math>\text{(A)}\,</math><dollar/><math>28000\qquad \text{(B)}\,</math><dollar/><math>32000\qquad \text{(C)}\,</math><dollar/><math>35000\qquad \text{(D)}\,</math><dollar/><math>42000\qquad \text{(E)}\,</math><dollar/><math>56000</math>
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<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>
  
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== Solution==
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==Solution 1==
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Let the income amount be denoted by <math>A</math>.
 
  
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We know that <math>\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}</math>.
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Let <math>A</math>, <math>T</math> be Kristin's annual income and the income tax total, respectively. Notice that
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<cmath>\begin{align*}
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T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\
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&= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\
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&= p\%\cdot A + 2\%\cdot(A - 28000)
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\end{align*}</cmath>
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We are also given that <cmath>T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A</cmath>
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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>
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Solve for <math>A</math> to obtain <math>A = \boxed{\textbf{(B) }\textdollar 32000} </math>.
  
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We can now try to solve for <math>A</math>:
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~ Nafer
  
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<math>(p+.25)A=28000p+Ap+2A-28000p-56000</math>
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==Video Solution by Daily Dose of Math==
  
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<math>.25A=2A-56000</math>
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https://youtu.be/io1JWEGl_J0?si=qa09mILGOdoNM5pc
  
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<math>A=32000</math>
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~Thesmartgreekmathdude
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So the answer is <math>\boxed{B}</math>
 
  
 
== 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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{{AMC12 box|year=2001|num-b=8|num-a=10}}
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{{AMC10 box|year=2001|num-b=8|num-a=10}}
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{{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 $p\%$ of the first $\textdollar 28000$ of annual income plus $(p + 2)\%$ of any amount above $\textdollar 28000$. Kristin noticed that the state income tax she paid amounted to $(p + 0.25)\%$ of her annual income. What was her annual income?

$\textbf{(A)}\,\textdollar 28000 \qquad \textbf{(B)}\,\textdollar 32000 \qquad \textbf{(C)}\,\textdollar 35000 \qquad \textbf{(D)}\,\textdollar 42000 \qquad \textbf{(E)}\,\textdollar 56000$

Solution 1

Let $A$, $T$ 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*} We are also given that \[T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A\] Thus, \[p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A\] \[2\%\cdot(A - 28000) = 0.25\%\cdot A\] Solve for $A$ to obtain $A = \boxed{\textbf{(B) }\textdollar 32000}$.

~ 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. AMC logo.png

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