Difference between revisions of "2004 AMC 12A Problems/Problem 5"

(Problem)
(Solution)
Line 7: Line 7:
  
 
==Solution==
 
==Solution==
It looks like it has a slope of <math>-\dfrac{1}{2}</math> and is shifted <math>\dfrac{4}{5}</math> up.
+
It looks like it has a slope of <math>-\dfrac{1}{2}<m<0</math> (looking at the intercepts) and is shifted <math>0<b<1</math> up. Try <math>m=-\dfrac{1}{3}</math> and <math>b=\dfrac{4}{5}</math>.
  
<math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math>
+
<math>\dfrac{4}{5}\cdot \dfrac{-1}{3}=\dfrac{-4}{15} \Rightarrow \mathrm {(B)}</math>
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2004|ab=A|num-b=4|num-a=6}}
 
{{AMC12 box|year=2004|ab=A|num-b=4|num-a=6}}

Revision as of 22:14, 3 February 2013

Problem

The graph of the line $y=mx+b$ is shown. Which of the following is true?

2004 AMC 12A Problem 5.png

$\mathrm {(A)} mb<-1 \qquad \mathrm {(B)} -1<mb<0 \qquad \mathrm {(C)} mb=0 \qquad \mathrm {(D)}$ $0<mb<1 \qquad \mathrm {(E)} mb>1$

Solution

It looks like it has a slope of $-\dfrac{1}{2}<m<0$ (looking at the intercepts) and is shifted $0<b<1$ up. Try $m=-\dfrac{1}{3}$ and $b=\dfrac{4}{5}$.

$\dfrac{4}{5}\cdot \dfrac{-1}{3}=\dfrac{-4}{15} \Rightarrow \mathrm {(B)}$

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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
Invalid username
Login to AoPS