Difference between revisions of "2001 AMC 12 Problems/Problem 3"
Brut3Forc3 (talk | contribs) (New page: == Problem == The state income tax where Kristin lives is levied at the rate of <math>p%</math> of the first <dollar/><math>28000</math> of annual income plus <math>(p + 2)%</math> of any ...) |
(→Solution) |
||
| Line 7: | Line 7: | ||
<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> | <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> | ||
| − | == Solution == | + | == Solution (unverified) == |
| + | Let the income amount be denoted by <math>A</math>. | ||
| + | We know that <math>\frac{A(p+.25)}{100}=\frac{28000p}{100}+\frac{(p+2)(A-28000)}{100}</math>. | ||
| + | |||
| + | We can now try to solve for <math>A</math>: | ||
| + | |||
| + | <math>(p+.25)A=28000p+Ap+2A-28000p-56000</math> | ||
| + | |||
| + | <math>.25A=2A-56000</math> | ||
| + | |||
| + | <math>A=32000</math> | ||
| + | |||
| + | 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}} | ||
Revision as of 17:42, 29 October 2008
Problem
The state income tax where Kristin lives is levied at the rate of $p%$ (Error compiling LaTeX. Unknown error_msg) of the first
<dollar/>
of annual income plus $(p + 2)%$ (Error compiling LaTeX. Unknown error_msg) of any amount above <dollar/>
. Kristin
noticed that the state income tax she paid amounted to $(p + 0.25)%$ (Error compiling LaTeX. Unknown error_msg) of her
annual income. What was her annual income?
<dollar/>
<dollar/>
<dollar/>
<dollar/>
<dollar/>
Solution (unverified)
Let the income amount be denoted by
.
We know that
.
We can now try to solve for
:
So the answer is
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 | |