Difference between revisions of "2001 AMC 12 Problems/Problem 9"

m (Problem)
m
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Let <math>f</math> be a function satisfying <math>f(xy) = \frac{f(x)}y</math> for all positive real numbers <math>x</math> and <math>y</math>, and <math>f(500) =3</math>. What is <math>f(600)</math>?
+
Let <math>f</math> be a function satisfying <math>f(xy) = \frac{f(x)}y</math> for all positive real numbers <math>x</math> and <math>y</math>. If <math>f(500) =3</math>, what is the value of <math>f(600)</math>?
  
 
<math>(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>
 
<math>(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
 
<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\mathrm{C}</math>.
 
<math>f(500\cdot\frac65) = \frac3{\frac65} = \frac52</math>, so the answer is <math>\mathrm{C}</math>.
 +
 +
== See Also ==
 +
 +
{{AMC12 box|year=2001|num-b=8|num-a=10}}

Revision as of 08:04, 15 February 2009

Problem

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$. If $f(500) =3$, what is the value of $f(600)$?

$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$

Solution

$f(500\cdot\frac65) = \frac3{\frac65} = \frac52$, so the answer is $\mathrm{C}$.

See Also

2001 AMC 12 (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 12 Problems and Solutions