Difference between revisions of "2006 AMC 8 Problems/Problem 4"

m (Solution)
(Solution 2 (Minor improvement))
Line 25: Line 25:
 
== Solution 2 (Minor improvement) ==
 
== Solution 2 (Minor improvement) ==
 
Note that full revolutions do not matter, so this is equivalent to going clockwise <math> \dfrac{1}{4}</math> revolutions and then counterclockwise <math> \dfrac{3}{4}</math> revolutions, making it ultimately go counterclockwise <math> \dfrac{1}{2} </math>, having the spinner point <math> \boxed{\textbf{(B)}\ \text{east}} </math>.
 
Note that full revolutions do not matter, so this is equivalent to going clockwise <math> \dfrac{1}{4}</math> revolutions and then counterclockwise <math> \dfrac{3}{4}</math> revolutions, making it ultimately go counterclockwise <math> \dfrac{1}{2} </math>, having the spinner point <math> \boxed{\textbf{(B)}\ \text{east}} </math>.
 +
 +
~JeffersonJ
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2006|num-b=3|num-a=5}}
 
{{AMC8 box|year=2006|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:38, 20 June 2022

Problem

Initially, a spinner points west. Chenille moves it clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions. In what direction does the spinner point after the two moves?

[asy]size(96); draw(circle((0,0),1),linewidth(1)); draw((0,0.75)--(0,1.25),linewidth(1)); draw((0,-0.75)--(0,-1.25),linewidth(1)); draw((0.75,0)--(1.25,0),linewidth(1)); draw((-0.75,0)--(-1.25,0),linewidth(1)); label("$N$",(0,1.25), N); label("$W$",(-1.25,0), W); label("$E$",(1.25,0), E); label("$S$",(0,-1.25), S); draw((0,0)--(-0.5,0),EndArrow);[/asy]

$\textbf{(A)}\ \text{north} \qquad  \textbf{(B)}\ \text{east} \qquad  \textbf{(C)}\ \text{south} \qquad  \textbf{(D)}\ \text{west} \qquad  \textbf{(E)}\ \text{northwest}$

Solution 1

If the spinner goes clockwise $2 \dfrac{1}{4}$ revolutions and then counterclockwise $3 \dfrac{3}{4}$ revolutions, it ultimately goes counterclockwise $1 \dfrac{1}{2}$ which brings the spinner pointing $\boxed{\textbf{(B)}\ \text{east}}$.

Solution 2 (Minor improvement)

Note that full revolutions do not matter, so this is equivalent to going clockwise $\dfrac{1}{4}$ revolutions and then counterclockwise $\dfrac{3}{4}$ revolutions, making it ultimately go counterclockwise $\dfrac{1}{2}$, having the spinner point $\boxed{\textbf{(B)}\ \text{east}}$.

~JeffersonJ

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png