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
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<math>\text{(A)}\,\textdollar 28000 \qquad \text{(B)}\,\textdollar 32000 \qquad \text{(C)}\,\textdollar 35000 \qquad \text{(D)}\,\textdollar 42000 \qquad \text{(E)}\,\textdollar 56000</math>
 
<math>\text{(A)}\,\textdollar 28000 \qquad \text{(B)}\,\textdollar 32000 \qquad \text{(C)}\,\textdollar 35000 \qquad \text{(D)}\,\textdollar 42000 \qquad \text{(E)}\,\textdollar 56000</math>
  
== Solution==
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==Solution 1==
 
 
=== Solution 1 ===
 
 
 
 
Let the income amount be denoted by <math>A</math>.
 
Let the income amount be denoted by <math>A</math>.
  
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So the answer is <math>\boxed{B}</math>
 
So the answer is <math>\boxed{B}</math>
  
=== Solution 2 ===
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 +
==Solution 2==
  
 
Let <math>A</math>, <math>T</math> be Kristin's annual income and the income tax total, respectively. Notice that
 
Let <math>A</math>, <math>T</math> be Kristin's annual income and the income tax total, respectively. Notice that
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&= p\%\cdot A + 2\%\cdot(A - 28000)
 
&= p\%\cdot A + 2\%\cdot(A - 28000)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
We are also given that <cmath>T = (p + 0.25)\%\cdot A = p\%\cot A + 0.25\%\cdot A</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>
Thus, <cmath>p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cot A + 0.25\%\cdot A</cmath> <cmath>2\%\cdot(A - 28000) = 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 = 32000</math>. <math>\boxed{B}</math>
 
Solve for <math>A</math> to obtain <math>A = 32000</math>. <math>\boxed{B}</math>
 +
 +
~ Nafer
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:07, 1 May 2021

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?

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

Solution 1

Let the income amount be denoted by $A$.

We know that $\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}$.

We can now try to solve for $A$:

$(p+.25)A=28000p+Ap+2A-28000p-56000$

$.25A=2A-56000$

$A=32000$

So the answer is $\boxed{B}$


Solution 2

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 $A = 32000$. $\boxed{B}$

~ Nafer

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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