Difference between revisions of "2001 AMC 12 Problems/Problem 3"

(Solution 2)
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=== Solution 2 ===
 
=== Solution 2 ===
  
Let <math>A</math> be Kristin's annual income. Notice that <math></math>A = p\%\cdot28000 + (p + 2)%\cdot(A - 28000) = [p%\cdot28000 + p%\cdot(A - 28000)] + 2%\cdot(A - 28000) =
+
Let <math>A</math> be Kristin's annual income. Notice that <cmath>A = p\%\cdot28000 + (p + 2)%\cdot(A - 28000) = [p%\cdot28000 + p%\cdot(A - 28000)] + 2%\cdot(A - 28000) =</cmath>
  
 
== See Also ==
 
== See Also ==

Revision as of 17:42, 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 $p\%$ 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

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 $A$ be Kristin's annual income. Notice that

\[A = p\%\cdot28000 + (p + 2)%\cdot(A - 28000) = [p%\cdot28000 + p%\cdot(A - 28000)] + 2%\cdot(A - 28000) =\] (Error compiling LaTeX. Unknown error_msg)

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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