Difference between revisions of "2001 AMC 12 Problems/Problem 3"
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=== Solution 2 === | === Solution 2 === | ||
− | Let <math>A</math> be Kristin's annual income. Notice that <cmath>p\%\cdot28000 + (p + 2)\%\cdot(A - 28000)</cmath> <cmath>= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000)</cmath> <cmath>= p\%\cdot A + 2\%\cdot(A - 28000)</cmath> | + | Let <math>A</math>, <math>T</math> be Kristin's annual income. Notice that |
+ | <cmath>\begin{align*} | ||
+ | p\%\cdot28000 + (p + 2)\%\cdot(A - 28000)</cmath> <cmath>= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000)</cmath> <cmath>= p\%\cdot A + 2\%\cdot(A - 28000)</cmath> | ||
+ | |||
+ | \begin{align*} | ||
+ | ((2x+3)^3)' &= 3(2x+3)^2 \cdot (2x+3)' \ | ||
+ | &= 3(2x+3)^2 \cdot 2 \ | ||
+ | &= 6(2x+3)^2. | ||
+ | \end{align*} | ||
== See Also == | == See Also == |
Revision as of 17:55, 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.
Contents
[hide]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
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. Notice that
\begin{align*} p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) (Error compiling LaTeX. Unknown error_msg)
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.